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CONCRETE 
PLAIN AND REINFORCED 


CONCRETE 


PLAIN AND REINFORCED 


VoL. IL. 
THEORY AND DESIGN OF CONCRETE 
AND REINFORCED STRUCTURES 


BY 
Tue Late FREDERICK W. TAYLOR, 
SANFORD E. THOMPSON, 5.B. 


AND 


EDWARD SMULSKI, C.E. 


With a Chapter by 
HENRY C. ROBBINS 


FOURTH EDITION 
TOTAL iSSUE, THIRTY THOUSAND 


NEW YORK 
TOHN WILEY’ & SONS, Ino. 
Lonpon: CHAPMAN & HALL, Limrrep 
1925 


CopyRIGHT, 1925 By 
E. W. CLARK 38rd 
AND 
SANFORD E. THOMPSON 


Copyrighted in Great Britain 


Printed a Lies mAs 


PRESS OF Et 
1/28 BRAUNWORTH & CO 
BOOK MANUFACTURERS 
BROOKLYN, N. Y, 


PREFACE TO FOURTH EDITION 


Tuis treatise is adapted for use both by the practical engineer 
and as a text-book for students of engineering. As a text-book it 
provides the student with material that is an essential part of his 
equipment for any line of professional engineering work. 

The Fourth Edition presents an entirely rewritten book in three 
volumes. The first volume treats of reinforced concrete design 
and construction; the second covers concrete materials and con- 
struction; and the third, special structures in reinforced concrete. 
Prefaces to Volumes II and III will be found at the beginning of 
these volumes. 

The authors, in preparing the work, have drawn upon their long 
and varied experience and practice as engineers. Because of this 
practical experience, they can speak with authority and can make 
specific recommendations with confidence. 

Reference to the Table of Contents indicates the breadth and 
scope of the treatment. The numerous illustrations are selected, 
not as pictures, but each with the definite purpose of conveying 
information on a particular point. The diagrams indicate the 
action of stresses. The line drawings show construction details 
based on the experience of the authors, to indicate the best way 
of attaining a required result. The photographs have been selected 
to show typical structures or structural details. The treatment 
of flat slabs, which the engineer—even one having broad experience 
in other types of construction—finds so difficult to design properly, 
is particularly comprehensive and practical. Not only are the 
theory and methods of the main design presented in simple and 
usable form, but all of the supplementary features, such as column 
heads, wall panels, openings, and beams, are taken up in practical 
fashion and with numerous illustrations. The treatment of column 
design includes the various types and adaptation to special uses and 

Vv 


Sf — 


vi PREFACE 


various City Building Codes. The portions dealing with founda- 
tions, footings, and piles cover all conditions. 

The treatment of building construction is particularly compre- 
hensive, covering not merely general structural features but also 
intimate consideration of various types of design, with special chap- 
ters on Wall Bearing Construction, Basement Walls, Roofs, Stair- 
ways, Fire Exits and Elevator Shafts, Steel Window Sash, Struc- 
tural Plans, Different Types of Buildings, and Chimneys. Two 
particularly interesting and novel chapters are those on the construc- 
tion in concrete of Theaters and Auditoriums and the Architectural 
Treatment of Exterior and Interior of Reinforced Concrete Build- , 
ings, the latter chapter prepared by Mr. Henry C. Robbins. 


SANFORD EH. THOMPSON, 


EDWARD SMULSKI. 
New York, June, 1925. 


CONTENTS 


CHAPTER I 


MATERIALS AND Mrtruops ror MAKING CONCRETE 


PAGE 
ET ei vy wilh hw OER se ee re we wa pe eee 1 
(EP ge toy y 2 =o a FE on = ae 2 
re ah ohn isk ris ein ap win ace as wa doef ale vide ee 2 
Coarse Aggregates............ PI WRI. Wer Pat Nee honk home Soe 3 
I re es eee en oS hes Hee ee ak EEL 4 
Tt ee eG re ea ee i ee a ee 4 
SEO CP ere BTACLCTISUICS «5 o5 erin ac ks ye cae ee ee CIN 5 

CHAPTER II 

REINFORCEMENT 
Metre remiorcne Miceli... ae Ey) s ks | tadib ee fee nile p yee tenn ee geen a 
Selection of Grade of Steel for Reinforcement.......... EET ONG OH ee 8 
Os MOSS nSG) oe 2) 001) fly oe rae rete eerta arr 10 

CHAPTER III 

Tests oF REINFORCED CONCRETE 

eT ieee ce cee Cle en ch ees ee A ee Le eS 35 
Tests of Beams to Determine Effect of Diagonal Tension................. 38 
behavior of Beam Failing by Diagonal Tension...............5.......... 46 
Pee without near Reinforcement... i... Gi. cna gene Sade a eee le 46 
Beams Reinforced for Tension and Compression...........-.--++-+++++-- 48 
fests ot Bond between Concrete and Steel. ......-0. 6.206262 e cece eee eds 51 
SEAS eae Lng. a ANS bh AP. TRAY Seley eee e Peres 51 
og. eM keyosrce orot | 2 S79 7 a ee ee i ra a gre rere rorare rier 56 
Splices of Tensile Reinforcement at Points of Maximum Stress............ 61 
Ree Tee ee enh eo oe eR OED, PS Oe 61 
Peer Otis GCAMS) 6c cce. cies da ee Coe wow ee pede oietwaal Pe 64 


vill CONTENTS 


PAGE 
Distribution of Concentrated Load on Wide Slabs.......----.+++++++s505 69 
Tests to Determine Distribution of Load by Slab to Joists............-++- 74 
Tests of. Plain Concrete Columns... .. 0. 2omn- 3-5 ++ < 06 =n eee a 76 
Tests of Columns Reinforced with Vertical Steel. ........-..------+-+05 78 
Tests of Spiral Columns... 2. . )26). 20 22 ey 21 ©) e seein ares ener 82 
Tests of Square Columns with Rectangular Bands... cds Se aAeee aegis coe 88 
Tests of Columns with Structural Steel Reinforcement.........--.+.+---: 88 
Tests of Long Columns... . 05.0. 062 pir >: +: + ee 93 
Torsional Resistance of Concrete and Reinforced Concrete te, sree eee 94 
Tests of Flat Slab Construction. ....0. 00... -5 + >>> > | =e 98 
Test of Flab Slab—Two-way System. .. 2.00... .)2 54+ +2) 5 ie seats 99 
Test of Flat Slab-—Smulski System. .....+..--/.-+++ 500% >= 73 ea 104 
Tests of Reinforced Concrete Buildings under Load . a. . us boeken 113 
Tests of Octagonal Cantilever Flat Slabs.......-.-++-++++2srerssrsseces 116 

CHAPTER IV 
THEORY OF REINFORCED CONCRETE 
General Principles of Reinforced Concrete Beams. ... +15 > 3: eens 123 
‘Assumptions.......--++2:e0s0ree sce t ett en ieee 126 
Analysis of Rectangular Beams. ......-----++++++ssrrttcrss ster stseees 126 
Notation: ....seseee + cece ewe tere emised ons ¢ scleen cyt: in 128 
Formulas for T-Beams.....4- 2-6) 00-4 29RE TEU «>: -e 133 
Reinforced Concrete Beams with Steel in Top and Bottom. sea 137 
Wedge-shaped Beams........... 22000 272.005 5s « 5 » vip sinter 140 
T-Beams with Unsymmetrical Flange..........-.+++++ssssesrtressctss 142 
Shearing Stresses in a Beam or Slab.......-+-----+sssr erst uses sees 143 
Diagonal Tension... 02s ety Sed re eee = «00 3 vee ea 147 
Column Formulas: :.... $$ 02542 foe ee ob te fos 8 + oe eS ee 158 
Members Subjected to Direct Tension......---+--+++ssserssr ests .. 162 
Notation... c.cccue ces ce ce ee oe memo ce esis sl 9) chan 167 
Plain Concrete Section under Direct Stress and Bending: /)5. eee 169 
Reinforced Concrete Section under Direct Stress and Flexuredss-2),. so 173 
Rectangular Section with Symmetrical Reinforcement.........------+++-: 174 
How to Combine Thrust with Bending Moment........------+++:s+++00: 187 
Members Subjected to Direct Tension and Bending. .:.; ..... 0:6 in een 189 
Flat Slab Theory... .....:0+> ++ sen nin eo: tenet ot) 90 eee a ee 194 
CHAPTER V 
REINFORCED CONCRETE DESIGN 
Ratio of Moduli of Elasticity. ......----+- sees e er ereceeeecees ss aan te oe 201 
Formulas for Design of Rectangular Beams.........+++++ererreresrcrres 203 
Design of Slabs... 20...) ccc.eee ce temer tess 00: + =) ai alee! 208 


Design of T-Beams.... 25.0004 0-525 + heroes 0's > =)s Ap tiie 215 


CONTENTS Ix 


PAGE 
eon Compression Reinforcement. ov iliac. vid Deda de Oe cee 230 
Rectangular Beam with Compression Reinforcement...............++.55. 232 
Examples of Beams with Steel in Top and Bottom...................... 240 
(ANSE TS ee ee 241 
Meemalamior Diagonal Tension... . 20... ee esi neem tees 247 
Semi-graphical Method of Spacing Stirrups............ 6... s seer eee 252 
Diagonal Tension for Moving Loads.............. 000. e eee eee eee 258 
NI BE ECCT OLE La Saleen souk ka i eo Rk te ed Le te alors 260 
Anchorage and Splicing of Reinforcement.........-... 600-5 seer eee ee. 265 
MET SCS CS cee oe eo de eee ele le ea BEGG Pea GG ea BAS 270 
Protective Covering for Reinforcement.............---2 see ee eee eee ees EE 
Clear Distance between Parallel Bars in a Beam..............----. 0005: 213 
Bending Moments for Use in Design of Beams...........----++.0++ +005. 275 
ee saa aa oe 277 
Formulas for Single Span Beams...... 6... -- sees eee eee teens 278 
Continuous Beams of Equal Spans...........-. ee eee eee teen ines 278 
Continuous Beams of Equal Spans Running into Columns................ 279 
Continuous Beams of Unequal Spans or with Non-uniform Loading........ 280 
Effect of Varying Moments of Inertia on Bending Moment.............-- 281 
ee are ON tIN0Us DSSTIS owe. eg ee ee eee 281 
Process of Designing Continuous Rene ee Ce ree eee 285 
Points at which Horizontal Reinforcement Should be Bent..............-. 290 
Reinforcement for Temperature and Shrinkage Stresses. .............---- 298 
CHAP IU VI 
‘Design or Fiat SLAB STRUCTURES 
Columns in Flat Slab Construction........ "ao eaairtd glieaie Ne Scobey Ae hg eeu py 305 
Interior Columns for Symmetrical Arrangement of Pandan, Oved ts Tatts 307 
ss SESH TIE ieee an hl en al il ie ae A ae al2 
eli LOAGS,.,...-.5+-+-+> Fee ti oe Dee eet Geter eoretee te Perea 319 
se ak fie lek hp Ry CMe ee ck oe we 322 
eee. e. s eb iies eoiee de ge er eee ee B20 
Bending Moments in Flat Slabs...........-:.-- 0s esse cree eects 328 
Bending Moments in Interior Panels in Flat Sai ewe stir eatae eet pros 331 
ey So ee ne ena area eo Cee a 303 
Vormuias for Slab Thickness... .....0525 00. (Gia Ca ae le ne aaa 336 
Thickness of Slab as Determined by Positive Bending Moment........... 340 
Compression Stresses in Concrete in Fat Sla bee ic ccs ler Am eae 341 
Shearing Stresses in Flat Slabs..........- +. eee cere eee eee renter ees 346 
Meetarcoment in Plat Slabs... . 6 col... ne ee NE, BS EE Ine 351 
NN ERR icc. ie ee ove wiog ee en ue RAMAN EL ek AES 358 
MEE ee caine nw 2 oe, Pans Shoe We eRe NER E RI oe Oh 358 
at er ire es bs pe ratansontem bh ace An MeN Be ESO A 362 


Circumferential or Smulski (S. M. I.) System.......------+--5- ptt oie IOOE 


x CONTENTS 


PAGE 
Explanation of Action of the Smulski System....................5+-505: 369 
Three-way. System . 1 oishssSeye ee Pa ener eer 374 
Openings in Flat Slab... 1.0% 3.50 SERRE. a IS Sk ee 375 
Design of Beams in Flat Slab Construction.....................+.++-+: 377 
Wall Beams. & £5 siwkeiin ebietecere sone ale ene lesecenpre= fo-/e GARG RUSRET Rs oeaiete enn 379 
General Remarks.:). ¢.cs- eo Se 385 
Problems in Designing Flat Slab Construction...............-.+-.5+555 389 
Example of Flat Slab Design............. PE 390 
Chicago and New York City Flat Slab Regulations............. Seok os 396 
Bending Moment Coefficients... 0.02.20. 00-52 See eee eens eet ees aas 398 

CHAPTER VII 
CONCRETE AND REINFORCED CONCRETE COLUMNS 
Plain Concrete Columns and Piers...2 ey oo. oe de 403 
Reinforced Concrete Colummns. . 454. oi 0. oe ae cin ees ote se 404 
Columns with Longitudinal Bars... 25.0220 4... «4; ©5040), eee 405 
Spiral Columns. o. 2... 0.0 000s cece ee ee ees oo» eel eo 419 
Octagonal Spiral Columns... . 6.000. oo ce ep «ou 0 arte 429 
Square Spiral Columns. ..¢% 2.6500. vee. ee oe oe 429 
Oblong Spiral Columns. - 2... 05. 0c net ee 2 2 eee 430 
Details of Spiral Columns’...... 062. 0.0. oc soe: oar 431 
Long Columms.:.. 2.2.5. Wi tes eens cone he ee 434 
Structural Steel Columns Imbedded in Concrete..................-4+-5-- 435 
Rules for Structural Steel and Cast-iron Columns................-.++.-- 435 
Rules for Structural Steel of Various Building Codes................-.-.- 437 
Details of Design of Steel Section. .............-.++- 5 ae= saya teste ony 440 
Economies in Column Design... 0... 2.5 -. «+ 0» +» aceite eae nee oon 445 
Reduction of: Live Loads in Buildings... ....... . . ... « . scjp one ene 452 
Design of Columns in a Building Several Stories High..............-+--. 456 
Columns Subjected to Bending... 03 6005500550 ce + | ween eee eee 458 
CHAPTER VIII 
FOUNDATIONS AND FOOTINGS 
Determination of Carrying Capacity of Soil................-.0 esse eens 469 
Bearing Power of Soils. 20 00.00 200000. 470 
General Rules of Design... ... 2.60 005. 0.0 ae. 472 
Plain Conerete Footings... 044 ...0c0-- 00h as 19 ee 480 
Reinforced Concrete Footings:. ..........-. 6605+. evs ea 481 
Independent Column Footings. «...............5-5 0. 42) i 485 
Design of Independent. Footings...................-5. 5+. =iaeninn ners 491 
Simple Slab Footing. ... 2.0... 200 cee eee tee ee es eer 503 


Sloped and Stepped Footings. ....27.0.........). 0 Joie 506 


CONTENTS xl 


PAGE 
Rectangular Footings... .......... cece eee e eee e eee e nee e tenet eeeteees 510 
Pe epePe tO OIE OOUITIES 2. oe oe ce ee ne ee tener reese wenaseers 514 
Independent Footings with Piles. ........ 1... eee seen eect eee e eee eee ees 516 
Continuous Wall Column Footings......... ay Pee JO concnton eee ee 518 
Pore POGUINES.. 6. hi ee eee eter e resent ester eeee 522 
Strap Beams to Connect Footings. ..........seeeeeeeeeeeeeereeeeeeeees 533 
Sg ee ee ee ene tee ee ecece Sa cee re ee ee pees) 

CHAPTER IX 

PILES 

Wood Piles corn t ens, facia elo so wa ee: he 6 88 0 0 8 0 eee eevee ee eeeoeeeneoe §42 
vw So a ee eee eo oe eee Bae ee ras ie «okcle OU 

CHAPTER. & 

Bumpinc ConsTRUCTION 

Relative Cost of Buildings of Different Materials.............. Sr eiie ie OU 
Actual Cost of Reinforced Concrete Buildings..............++eeees B eteare 567 
a eh. nll Pel ene eee wae tees ewes ee’ 569 
Meme ra DCSOTIUIOW es sic. sc oe ee te ete treo eneeeees 572 
Beam and Girder Design of Floors. ..........-- eee c eee eee e eee eens 575 
Example of Beam and Girder Design. ..........-- ++ esse tere eeees cere 578 
Tint elab Floor Construction. . 6.2.22. 6 eee ee ee ee tee eee eens 587 
Light-weight Floor Construction. ........+ ++ esses er ret eeceeeerete te 588 
Reinforced Concrete Hollow Tile Floor Construction............+++e++++: 589 
Weumple of Hollow Tile Floor. ..-........-.-+2 see ree cece een tne rene: 598 
SI aD cary Se  e te  e es ee ee ee ee we eu se 602 
Combination of Structural Steel and Concrete. .......--. seer cere cece 611 
Basement and Ground Floor Slabs... 2.0.0... 66sec teeter eee eee enees 618 
Pg ecb nse se enue see Met te bee ee tases 620 

CHAPTER XI 

WALL-BEARING CONSTRUCTION 

Brickwork.5.......»- Ce CURR CUR FO he Soa it gine nine feck we 629 
“Coppa: WCU eae e Bee Ae os Ct er che ened ee 634 


CHAPTER XII 
BASEMENT WALLS 


BMPR OT Construction... fs. c ee ee es toes oce mond tame eines 637 
auch TE SREY el ai a a ee ee a a) or 638 


Xl CONTENTS 


; PAGE 
Wall Supported at Top and Bottom.............. 2 es 638 
Wall Supported by Colunttiges 2.2) on oeeeete eee pee oe yt a ee 643 
Example of Basement Wall Supported by Column....................... 645 

CHAPTER XIII 

Roor CONSTRUCTION 

LORIN Goss 2 degas e heee. - Seewh ne so in: ep va guednte 9 2 9 ck re ee 648 
Drainage... 0.0.0 cers we bo a eee ue ote nine eae + see nen 649 
Insulation... 5 jahge ee ee ne RR 2 2 650 
Roof Coverings. oui 2s. «+ os. 60 ep ee Ge Se oe as oe ce 654 
Roof Désign’ (te 0 08 og seco ae a ' 659 
Long Span Rosf Construction.; . 20%. es. See ee 661 
Concrete Arches for Long-span Roofs... <2. $05 52 2a ei ee 668 
Long-span Roof Construction. .. 4. deeply gee ss <i 
Sawtooth Roofs? ... .. 20.6% ees nina es as Seg see eee ob One ee 675 
Pre-cast Concrete Roof Trusses is so sacks? .Sategia..) 0... ee 677 
Auxiliary Structures above Roof... 2... 0... +25 5i- 509 ee en 678 

CHAPTER XIV 

Srarrways, Fire Exits, AND ELEVATOR SHAFTS 

Prromaxite uc bo". 5s ere ey OU OA, So 682 
Layout'of Stairs... 20008 2 686 
Structural Design of Staira: . 2.0... 02.) 220) 690 
Spécial Design of Stairs... er 695 
Elevator Shafts... .. .. BA a rr Hb weet 695 

CHAPTER XV 

STrEL WINDOW SasH 

Standards of Window Sash... ... . occ aah so = oe one 699 
Ereetion of Sash... ct cece ec ce cee ccs eee oe 6 oo) + ee 702 
Sizes of Sash... ... .. «+ « «mabrh wos wate 2 mb con gen koe er 705 
Underwriters’ Prvoted Sagiits tke ea eee ee Pe ete a 706 

CHAPTER XVI 

STRUCTURAL PLANS FOR BUILDINGS 

General Structural Plans... .siiecl)- 4h este whi ee eee 707 
Complete General Structural Plans.........-... .. .. =] gee 710 
Method of Showing Details... .7 2224 52-20. 0... 0. 1s 2 716 


Working Plans :...% 402 ee PP 720 


CONTENTS Xl 


CHAPTER XVII 


ARCHITECTURAL TREATMENT OF EXTERIOR AND INTERIOR OF REINFORCED 
CoNCRETE BUILDINGS 


By Henry C. Robbins 


PAGE 
Re ee, ges oes os Gn 4d ds oad wily 0a Dae oll lb eleareks 729 
Concrete in Combination with Brick, Tile or Other Masonry.............. 739 
I Th vie Sas ek on 2 98 Le wee de s wR ad bey S 747 
Renee GUS SOXEETION «666s. 6 sa sls vild a TR cape spa tall fae been L dee 753 
REG cr sk eR Ga nck so pwd 6 a wav He ous 6 Gawd 155 

CHAPTER XVIII 
CONCRETE IN CONSTRUCTION OF THEATERS AND AUDITORIUMS 
TG > a eg ec 762 
eI Ny Fy nk ce ky wos cee kk wh Rage eg eee Ae 763 
ee ER RIS op a a 765 
(FRE N25, oe RM aS a a 772 
em et 1 OS EVUG IOI G6 ig whe beck es hehe peck eka ve eae wee et es 778 
Spe ey SGU M me AIGIN a) yc hn ph eile ends os Ge Wks ee en ve ewe 778 
CHAPTER XIX 
REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 
Warehouse Construction....... at Oe ee REPRE Se Xr AAI Stee 785 
(UE ALG Te UTP ae a ON ay aa Sr eae 792 
I eeCNIOR ro oes 5 oc ht eee eeews wee rw aa es) ee 792 
Memriotaidimes and Garages... 00. Se ee bp eee ees 795 
ET: remem er gs) 8a Ye pall vie oes Be 6 a AOU ee ee es 803 
Perelen srervicnt Houses and Hospitals... 22. 6... de ns bee cee cece 805 
NEG yy eee ve ee ey tea ta Cries Me ONE ota 808 
CHAPTER XX 
REINFORCED CONCRETE CHIMNEYS 

UTI ee RASS Soo. Tt se dae De vedas Seomw Golvgepie Ce anne) Ames 812 
Theory and Formulas........... Pa Ee eee ere ENS PU eral: 813 
meer rioow Circular Beams, .... 2.650 ce ete ewe eed ey ww ee 822 
ieee: meintorcea Concrete Chimneys... /...-........ 536.0 0e gree an nee 822 


eer erate amtie vol esien eG. ks SUS We ieee ee we dels 829 


xiv CONTENTS 


CHAPTER XXI 


RETAINING WALLS 


PAGE 
Weight of Barth... 2.242. -- pbk be ce ered's bbe bt eet tulad an 833 
Earth Pressure... .. . . ... <5 cc fame Bee 6e oh eee 0 ptt a ete anata ae 835 
Design of Retaining Walls of Gravity Section............+...++sesseeeee 839 
Cantilever Walls... ss. 5. co ub bb oR te teed be pe 849 
Example of T-shaped Retaining Wall................++...45. ee ree 856 
Wall with Counterforts: ..... 0b. ..554.065%9 028. 0.) Sa 863 
Special Designs of Retaining Walls.............-. 5-6. s esses eres tee aee 869 

CHAPTER XXII 
TABLES AND DIAGRAMS 
Creneral. oc; cw ec 0 ot bejask enue.» ot lcel aleene steels Bienen ante he nn 879 
Constants for Rectangular Beams and Slabs.............---++++++s05-:- 880 
Safe Loading for Rectangular Beams............... 2. esses eeeere ee eees 884 
Safe Loading for Slabs............ 0002500002051 00 +00 >= 9 gO ee 886 
Design of T-Beams..... 00... 000.0 0 Rev ee oe 2 + > =o 5 ose 894 
Design of Stirrups........0.-.00+-205 +++ e se ++ + gee 899 
Beams with Steel in Top and Bottom............... 2. se sees eee eee eee 904 
Flat-Slab Design... 25.0000 0. 85022 oe cee ny oetnt ne 911 
Column Design... .....-.60. 08 s2 > seh ai ep + 2 oe ee 915 
Members Subjected to Direct Loading and Bending..............------- 934 
General Data... occ. ... 0 cue Sone eyeqei en eae + one +e 2 eer 942 


CONCRETE 
PLAIN AND REINFORCED 


CHAPTER I 
MATERIALS AND METHODS FOR MAKING CONCRETE 


___In this chapter are treated the materials used in concrete and 


ERRATUM 


Page v (Preface).—Lines 7 and 8, beginning with the semicolon 
on line 7, should read as follows: the second volume will cover 
Theory and Design Formulas for Continuous Beams, Frames, 
Building Frames, and Arches; while the third will treat of Con- 
crete Materials and Construction. 


ESSENTIALS OF GOOD CONCRETE 


Concrete consists of hard particles of inorganic material, called 
aggregates, such as sand, gravel, crushed stone, slag, or cinders, 
cemented together with Portland cement and water. When all the 
ageregate is fine, like sand, the mixture is termed mortar. 

In making durable concrete, the prime essentials are as follows: 


Hard, durable aggregates; 

Aggregates graded, or stepped, in size, from fine to coarse; 

Aggregates clean, that is, entirely free from organic impurities; 

Cement sufficient in quantity to give the required strength 
or water-tightness ; 

Water clean and free from organic matter or deleterious 
minerals; 

Quantity of water to produce the consistency needed for the 
work, always avoiding a wet, sloppy mix; 


XIV CONTENTS 


CHAPTER XXI 


RETAINING WALLS 


Weight of Earth........... PH CRE ete 2, « 
Earth Pressure... «+ o:+: {vic AOC te set Sh erent, rae ere 
Design of Retaining Walls of Gravity Section............... 
Gantilever Walls: . .<.¢.sc et te ie ie te 
Example of T-shaped Retaining Wall................-+.++. 
Wall with Counterforts <2 5.44 520: 0.50. 520% se 2 2 
Special Designs of Retaining Walls...........--..--+++++++ 


CHAPTER XXII 


S, ¢) ohm \w) Bid. 0 6),8 Pe 


o © @ 8 « 6) © 6, @ @ ate 


CONCRETE 
PLAIN AND REINFORCED 


CHAPTER I 
MATERIALS AND METHODS FOR MAKING CONCRETE 


In this chapter are treated the materials used in concrete and 
certain fundamental principles relating to concrete. The properties 
of concrete are fully discussed in Volume II of this treatise, which 
covers selection and testing of materials, proportioning, strength, 
water-tightness, resistance to sea water, fire resistance, effect of 
impurities, and the various other characteristics, as well as dis- 
cussion and advice on mixing and placing concrete, form construc- 
tion, architectural treatment of surfaces, and the laying of floors, 
roads, and pavements. In this chapter, only such facts and con- 
clusions are given as are necessary for the understanding of the 
subject matter of this volume. 


ESSENTIALS OF GOOD CONCRETE 


Concrete consists of hard particles of inorganic material, called 
aggregates, such as sand, gravel, crushed stone, slag, or cinders, 
cemented together with Portland cement and water. When all the 
aggregate is fine, like sand, the mixture is termed mortar. 

In making durable concrete, the prime essentials are as follows: 


Hard, durable aggregates; 

Aggregates graded, or stepped, in size, from fine to coarse; 

Aggregates clean, that is, entirely free from organic impurities; 

Cement sufficient in quantity to give the required strength 
or water-tightness; 

Water clean and free from organic matter or deleterious 
minerals; 

Quantity of water to produce the consistency needed for the 
work, always avoiding a wet, sloppy mix; 


2 MATERIALS AND METHODS FOR MAKING CONCRETE 


Mixing of the mass thoroughly to insure homogeneity and 
also properly to work the cement; 

Transporting and placing in such a way as to avoid separation 
of aggregates; 

Ramming or puddling so that the concrete will ail every part 
of the forms; 

Temperature of concrete maintained appreciably above 
freezing until it is thoroughly hard, to avoid retarded 
hardening. 


CEMENT 


Portland cement is used universally for reinforced concrete and 

also for plain concrete, except where abnormal conditions necessitate 
a special cement. 

Cement should be tested, except fon very small and unimportant 
jobs, and should pass the requirements of the American Society for 
Testing Materials for tensile strength in 1:3 mortar; fineness; 
setting time; and soundness. Samples of cement, except for small 
jobs, are preferably taken at the mill before shipment. If this is 
impracticable, about 20 bags out of a carload of, say, 200 barrels 
(800 bags) are opened, and a small quantity taken from each and 
mixed together to make a single sample of about 2 quarts. 

Cement that is unsound, that is, cement that checks or warps 
when steaming, should not be used. Sometimes the unsoundness 
disappears if the cement is held for a period. 

A particularly dangerous characteristic in cement is a “ flash set,” 
which is evidenced in practice by the concrete setting so quickly as 
to harden in the mixer or in ordinary transportation to place. Con- 
crete thus prematurely hardened is practically worthless and must 
not be reworked. 


FINE AGGREGATES 


Poor sand is the most common cause of defective concrete. 
The presence of organic matter, such as vegetable loam, is the most 
objectionable characteristic, as such matter, even when present in 
very small quantities, prevents the concrete from hardening satis- 
factorily. Concrete containing organic matter may remain soft and 
friable for months and never attain full strength or resistance to 
abrasion. 


AGGREGATES 3 


Sand, Therefore, Should Always be Tested.—It is more neces- 
sary to test the sand than the cement, as the quality of sand cannot 
be determined by inspection. With the exception of the presence of 
organic matter, the quality of a sand that is most detrimental to 
concrete is excessive fineness. Not only must an excess of fine 
particles of dust be avoided, but the average size of the grains should 
be coarse. 

Sand, for most uses, is preferably graded from fine to coarse, 
with not more than 30 per cent passing through a 50-mesh sieve and 
not less than 85 per cent passing through a No. 4 sieve. 

The most important test of sand is for organic matter. The 
Abrams-Harder test is described in Volume II of this treatise. 

Crushed stone screenings may be used as a substitute for sand or 
may be mixed with sand, provided they do not contain an excess of 
dust, which tends to come to the surface of the concrete and reduce 
its durability. Limestone screenings, for this reason, make a very 
poor material for roads and sidewalks. 


COARSE AGGREGATES 


Coarse aggregate should consist of crushed stone, gravel, or other 
approved inert materials with similar characteristics, or combina- 
tions thereof, having clean, hard, strong, durable, uncoated particles 
free from injurious amounts of soft, friable, thin, elongated or lami- 
nated pieces, alkali, organic or other deleterious matter. 

Coarse aggregate should range in size from fine to coarse within the 
limits given in table below. | 

The following table indicates desirable gradings for coarse aggre- 
gates of certain nominal maximum sizes: 





| 
Nominal Percentage by Weight Passing through Percentage Passing, 





Maximum Standard Sieves with Square Openings not More than 
Size of 
Aggregate, 
in. 3in. | 2in. |13in.| lin. | Zin. | }in. |No. 4 Sieve|No. 8 Sieve 
3 ND Elbe. oe BU (Sieh eel ip cas eat 10 5 
JEP SOT RA ONS Te asaic, UY (9 Da aig 1 POE Fee 10 5 
rg ae ae oa ee ee 40-75} 8.4 ss 10 5 
Lo SR con i emer Ree we ee oe ee 10 5 
o,f game ce Slee gee IR ONE BO ae 10 5 
Re ate Li A oe oust » Sif aals o's 95 10 5 





4 MATERIALS AND METHODS FOR MAKING CONCRETE 


The test for size and grading of aggregate should be made in 
accordance with the Standard Method of Test for Sieve Analysis of 
Aggregates for Concrete. (A.S. T. M. C 21-24). 


PROPORTIONS OF CONCRETE 


The amount of cement to use in concrete, that is, its proportion 
to the aggregates, depends upon tke nature of the structure and the 
strength or water-tightness required. The proportions are usually 
expressed in terms of cement, fine aggregate, and coarse aggregate; 
proportions 1 : 2 : 4, for example, meaning one part cement by volume 
to two parts of sand or other fine aggregate, to four parts of gravel 
or broken stone or other coarse aggregate. A bag (94 lb.) of cement 
is always assumed to be 1 cubic foot. A 1:2:4 mix, therefore, 
means 1 bag cement, 2 cubic feet fine aggregate, and 4 cubic feet 
coarse aggregate. 

Proportions in practice are apt to run from 1:1:2, often used 
in columns, to 1 : 4 : 7 for mass concrete employing specially graded 
aggregates. The usual proportions for reinforced concrete are 
1li'22 A 

For concrete in large masses, such as dams, a great deal of money 
can be saved by making extensive tests and special measurements of 
materials when laying, so as to Regs the maximum density and the 
minimum of cement. 


MIXING AND PLACING 


Concrete mixers, driven by power, are used almost universally 
for all kinds of concreting. ‘The size and type of machine must be 
adapted to the structure and to the quantity of concrete to be placed: 
ina day. For ordinary work, a revolving drum with interior blades 
is used. 

The time of mixing is an important element. The authors advise 
mixing a batch of concrete not less than { minute after the dry mate- 
rials and water are in the mixer. 

After the proportions have been fixed, the most important ele- 
ment in good mixing is the amount of water to use. The quantity of 
water required for a batch, with the materials as they generally come, 
should be determined by trial with the aid of the slump test. The 
water for every batch should be measured. When the standard 
quantity has been fixed, the volume may be changed slightly to 


STRENGTH AND OTHER CHARACTERISTICS 5 


allow for difference in moisture of sand or for the working of the 
concrete in the mass, since as the depth of the concrete increases, 
the upper part of the mass becomes wetter. 

Concrete is often seriously injured by an excess of water in 
mixing. The strength is reduced by excess water, the wearing 
qualities are lessened, danger of checking or crazing of the surface is 
increased and laitance, a whitish, nearly inert substance, rises to the 
surface. For reinforced concrete, the best consistency is that which 
will flow very sluggishly without flattening out and without separa- 
tion of the fine and coarse material. If the concrete is deposited 
through a sloping chute, the slope should not be flatter than one 
vertical to three horizontal. For many mixtures, a one to two 
slope should be maintained. The criterion in fixing the maximum 
amount of water in concrete is (1) to avoid separation of cement and 
fine aggregate from the coarse aggregate, and (2) to prevent excess 
water or thin mortar from rising to the top of the mass as it is being 
placed. 7 

While the greatest danger in mixing concrete is that of having 
the mix too wet, excessive dryness also must be avoided. In rein- 
forced concrete, the mix must be soft enough to thoroughly imbed 
the steel and fill all parts of the forms. In mass concrete, a drier 
mixture can be used than in reinforced concrete. Enough water 
must be added to make the mortar flush slightly to the surface and 
to fill all voids. 

When, for small pieces of work, concrete is mixed by hand, the 
same precautions must be taken as in machine-mixed concrete. 
Materials must be carefully measured, a bag of cement being assumed 
to be one cubic foot; the materials must be spread out in uniform lay- 
ers; they must be thoroughly mixed dry by turning two or three times 
with a square-pointed shovel; water must be added to obtain the 
required consistency; the mass must be turned about four times 
wet, and the mixture taken to place without delay. 

In important construction, specimens of the concrete should be 
taken at intervals, molded in 6 in. by 12-in. cylinders to be crushed 
at the age of 7, 14, or 28 days. 


STRENGTH AND OTHER CHARACTERISTICS 


For reinforced concrete structures, a compressive strength of the 
concrete of 2000 Ib. per sq. in. at the age of 28 days is a common 


6 MATERIALS AND METHODS FOR MAKING CONCRETE 


requirement. The strength is determined by testing cylinders 6 in. 
in diameter by 12 in. high. 

For concrete in large masses under low stress, and for such struc- 
tures as cellar walls, a strength of 1 500 lb. per sq. in. at 28 days is 
satisfactory; while for columns, where small dimensions are pre- 
ferred, 3.000 lb. per sq. in. may be specified. The variations in 
strength to accord with any requirements are obtained by proper 
proportioning of the cement and grading of the aggregates. 

For water-tight work, concrete 2 000 lb. per sq. in. at 28 days is 
usually required. In work of this kind, the size of the voids, as well 
as the density of the concrete, is important; therefore, a finer sand 
may sometimes be advantageously used to make the concrete flow 
smoothly and to avoid stone pockets due to separation of materials. 

Concrete laid in sea water, or in alkali waters, requires special 
precautions, as discussed in Volume II. 

For fire resistance, reinforced concrete is the best all-round build-- 
ing material. Aggregates like trap are better than gravel, which 
has a large coefficient of expansion. or limestone, which is liable to be | 
affected chemically by the heat. 

For the purpose of design, the following strengths of 6 by 12 
cylinders at 28 days are assumed for various proportions of concrete: 


Strength of Cylinder at 28 days. Lb. per sq. in. 








Mix ye es Ba 17:13:38 1:24 2F25 
Strength at 28 days, fc’. 3 000 2 500 2 000 1 600 
n 10 12 15 15 


eee 


n=ratio of moduli of elasticity. 


CHAPTER II 
REINFORCEMENT 


Reinforcement for concrete may consist of any kind of ductile 
metal, the prime factors being strength, reliability, and economy. 
In the United States, steel is used practically exclusively, while in 
Europe wrought iron is used to a great extent. 


QUALITY OF REINFORCING STEEL 
The Specifications of the American Society for Testing Materials, 
endorsed by the authors, require the following properties for rein- 


forcement. 
Chemical Properties.—The steel shall conform to the following 


requirements as to chemical composition: 


Been havtis [ Bessemer not over 0.10 per cent. 
. | Open-hearth not over 0.05 per cent. 


Physical Properties.—The bars shall conform to the require- 
ments as to tensile properties given in the table below. 


Tensile Properties of Concrete Reinforcement Bars 


Plain Bars Deformed Bars 
: Rg et eer 5 ena as a aR eee 
Properties wal d 
Considered Structural] Inter- Structural] Inter- 
; Hard ; Hard Bars 
steel mediate den steel mediate eae 
grade grade z grade grade £ 








Tensile strength,} 55 000 70 000 80 000 55 000 70 000 80 000 | Recorded 








Ib. per sq. in.... to to min. to to min. only 
70000 | 85000 70 000 85 000 
Yield point, min., 
Ib. per sq. in....| 33 000 40 000 50 000 33 000 40 000 50 000 55 000 
Elongationin 8in., 


min., per cent. .}1 400 000*|1 300 000*|1 200 000*|1 250 000*/1 125 000*| 1 000 000* 5 


it ae SS) |S ae al ieee acereeaee (haar eae ama 


Tens. str. | Tens. str. | Tens. str. | Tens. str. | Tens. str. Tens. str. 


* Deduct 1 per cent for each increase of $-inch above 3-inch diameter, or for each decrease 
of zs-inch below ;4-inch diameter. 


8 REINFORCENEMT 


Bend-test Requirements.—The test specimen shall bend cold 
around a pin without cracking on the outside of the bent portion, as 


follows: 
Bend-Test Requirements 


Se Ee 




















Plain Bars Deformed Bars 
Thickness or Cold- 
ere Structural| Inter- Structural| Inter- wwisiee 
of Bar : ; Hard ; Hard Bars 
steel mediate eae steel mediate ete 
grade grade & grade grade = 
Under 7 in....:.- 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. | 180 deg. 
=f Gh he = ob a=t @ = ror ad =At a = 2t 
$ in. or over....- 180 deg. | 90 deg. 90 deg. 90 deg. 90 deg. 90 deg. | 180 deg. 
d— ese ak Greet O—so) de—war d = 3t 





Exprianatory Note: d = the diameter of pin about which the specimen is bent; 
t = the thickness or diameter of the specimen. 


The bending test is the most important test in the specification, 
and no steel which fails in this test should be used under any cir- 
cumstances. 

Steel with high elastic limit, whether due to high carbon or to 
manipulation in manufacture, should be purchased with these 
reservations even if the working stress is to be no higher than is 
used with structural grade steel, say, 16 000 lb. per sq. in., because 
it is liable to be brittle. In case a lot of steel has been delivered 
without previous test by the purchaser, one bar selected at random 
in every 100 should be subjected to this test; and if it fails to pass, 
the portion from which it is taken should be rejected. On small 
jobs, where it is impossible to get proper tests of steel, it is of impor- 
tance to purchase material from a reliable dealer having direct 
connection with reliable mills. Otherwise there is a likelihood that 
material rejected on some other job may be furnished. 


SELECTION OF GRADE OF STEEL FOR REINFORCEMENT 


Steel of Structural-steel Grade has been universally accepted 
in the past as the most satisfactory material for reinforcement. It 
‘5 manufactured under such standard conditions that, for unim- 
portant structures, it may be used without other test than the bend- 
ing test given above. The allowable stress for this grade should 
not exceed 16 000 Ib. per sq. in. This gives a factor of safety of a 
little over 2, for the live and dead load. The factor of safety for 


SELECTION OF GRADE OF STEEL FOR REINFORCEMENT 9 


the live load alone will depend upon the ratio of the live to dead 
load. For structures subjected to severe weather conditions or to 
locomotive smoke; it may be advisable to reduce the stresses so as 
to prevent formation of open cracks. 

Intermediate Grade Steel has the advantage over structural- 
steel grade in its higher elastic limit. It is possible to use a stress 
in steel of 18000 lb. and still have a somewhat higher factor of 
safety than for the structural-steel grade. Therefore, if inter- 
mediate grade steel can be bought at a unit price not exceeding that 
of structural-steel grade by more than 8 to 10 per cent, it may be 
used with economy. Since actually there is no appreciable differ- 
ence in cost between the two grades, the intermediate grade steel is 
gradually replacing the structural-steel grade. In some cases, 
intermediate grade is specified even if the allowable stress in steel 
is only 16 000 lb. per sq.in. The factor of safety is thereby increased 
considerably by the additional strength of intermediate steel, with- 
out any appreciable extra cost. Intermediate grade steel should 
be properly tested. 

The only disadvantage in using intermediate grade steel with 
18 000-Ib. stress is that the cracking of the concrete at working 
loads is larger, because the stretch in the steel, being proportional 
to the stress, is larger. Under ordinary conditions for ordinary 
building construction work, this objection is not important. Tests 
made in Europe (see Vol. IJ) have proved conclusively that the 
cement protects the steel from ordinary corrosive action until the 
elastic limit of steel is nearly reached. Where it is desirable, how- 
ever, for any reason, to avoid cracking, the stress in steel should be 
kept as low as for structural grade steel. 

For temperature reinforcement, a steel of high elastic limit is 
particularly valuable. 

Hard Grade Steel.—If properly tested as required by the pre- 
viously given specifications, hard grade steel may be safely used 
for reinforced concrete. Although hard grade steel has a higher 
elastic limit than the intermediate grade, the allowable stress should 
not be larger than 18 000 lb. per sq. in. because an accidental 
increase in loading would produce excessive cracking. With a 
stress of 18 000 Ib. per sq. in., the factor of safety for hard grade 
steel is larger than for either of the two previously described grades. 
What has been said as to the economy of using intermediate grade 
steel applies also the use of hard grade steel. 


10 REINFORCEMENT 


Many engineers do not approve of the use of hard grade steel 
because of its brittleness when of poor quality, and because it is pro- 
hibited in ordinary structural steel work. Brittleness in reinforcing 
steel, however, is less dangerous in reinforced concrete than in 
many classes of structural steel work because the concrete protects 
the steel from shock, and also because smaller sections of steel 
are used in concrete beams than in steel beams. The large and 
irregular shapes in structural steel render it much more sensitive 
to irregular cooling during the process of manufacture than is the 
case with reinforcing bars. 


Syapes USED FOR REINFORCEMENT 


Small Angles.—Small angles are sometimes used as reinforce- 
ment, mostly in columns or in Melan arches. 

Reinforcing Bars.—The most common shapes used for rein- 
forcement are round and square bars, and to a much smaller extent, 
flat bars. ‘The bars may be either plain or deformed. The com- 
mon types of deformed bars are shown in Fig. 1, p. 11. 

Plain vs. Deformed Bars.—Steel bars used in reinforced concrete 
may be either plain or deformed. 

Plain bars are round, square, or flat bars passed through regular 
rolls. . 

Deformed bars are bars provided with irregularities of the surface. 
formed either in the process of rolling or by twisting after rolling. 
When the area of cross section of deformed bars is not constant, its 
effective area is equal to the minimum cross section. 

The purpose of the deformations of the surface is to increase 
the bond between the bars and the concrete, by adding to the sur- 
face bond of the plain bars the mechanical bond of the projecting . 
deformations. ‘Tests show that deformed bars of proper design 
have a considerably larger bond resistance than plain bars, so that 
larger unit bond stresses are allowable. (See p. 54.) 

Either plain or deformed bars may be used with safety, provided 
the unit bond stresses are kept within the limits set for the two 
types of bars. In ordinary beams and slabs, the bond stresses are 
likely to be low, so that there is no difficulty in keeping the bond 
stresses within working limits of plain bars. In such cases, there is 
no special advantage in using deformed bars. In footings and short 
beams carrying heavy loads, on the other hand, it is often impos- 


TYPES OF REINFORCING BARS 





Cold Twisted Square Bar 























































































































































































































































































































































































































Herringbone Bar Monotype Bar 
Fic. 1.—Types of Deformed Reinforcing Bars. (See p. 10.) 


11 


“ 


12 REINFORCEMENT 


sible to utilize the full working value of the steel in tension, owing 
to the difficulty of providing sufficient bond. The use of deformed 
bars in such members is therefore advantageous and economical. 
They are also useful for temperature reinforcement, as the distri- 
bution of cracks depends upon the bond between the bars and the 
concrete. In road construction, deformed bars are useful for the 
same reason. 

As a general proposition, where deformed bars are easily obtain- 
able and their cost is not appreciably higher than that of plain 
bars, they are recommended in preference to plain bars because 
the factor of safety against actual collapse is larger for deformed 
bars than for plain bars. The use of deformed bars is particularly 
recommended for structures subjected to dynamic loads. 

Properties of Bars Used as Reinforcement.—The table below 
gives the sizes of plain bars used in practice, their areas, weights, 
and perimeters. 


Round and Square Bars 


Areas, Weight and Perimeters 


Se ee es 





Round Bars | Square Bars 
Site, 4) ot ae pees ake 2 ee 
inch Area, |Weight per | Perimeter, Area, |Weight per | Perimeter, 
sq. in. foot, lb. in. sq. In. foot, lb. in. 

1/4 0.049 0.17 0.78 0.062 0.21 1.00 
5/16 0.077 0.26 0.98 0.098 0.33 1.25 
3/8 0.110 0.38 1.18 0.141 0.48 1.50 
7/16 0.150 0.51 76 0.191 0.65 1.75 
1/2 0.196 0.67 1.57 0.250" 0.85 2.00 
9/16 0.248 0.84 Eger 0.316 1.08 2.25 
5/8 0.307 1.04 1.96 0.391 1.33 2.50 
11/16 0.371 1.26 2.16 0.473 1.61 2.75 
3/4 0.442 1.50 2.36 0.563 1.91 3.00 
13/16 0.518 1.76 2.55 0.660 2.24 3.25 
7/8 0.601 2.04 2.75 0.766 2.60 3.50 
15/16 0.690 2.35 2.95 0.879 2.99 3.75 
1 0.785 2.67 3.19 1.000 * 3.40 4.00 

11/8 0.994 3.38 3.53 1.266 4.30 4.25 5 


TYPES OF REINFORCING BARS 13 


The standard sizes in the table are set in bold type. The other 
sizes can be obtained with difficulty, especially when deformed bars 
are desired. Therefore, the designer is cautioned against using 
them. 

Cold Twisted Bars have the same dimensions and weight as the 
corresponding square bars. 

Corrugated Bars, round and square, have the same dimensions as 
the corresponding round or square plain bars. Their weight is 
somewhat greater, however, on account of the projections, which 
are not included in the effective area. 

Kahn Trussed Bars have special dimensions as given in table 
below. 


Dimensions, in. 








Weight per 
Tome oce bigs i | SCLineal Poot, Area, sq. in. 
D B lb. 
plist es a wis as 3 13 1.4 0.41 
Ay eee OS ark: a 2,3, yews 0.79 
I te SS oe ado ti Ss i; 24 4.8 1.41 
PEE Ee ss the ie 23 6.8 2.00 
By Wes Settee oe 66! ona: 2g os 2 34 10.2 3.00 











If trussed bars are used as beam reinforcement, the wings are 
sheared near the ends and bent up to serve as diagonal tension 
reinforcement. 

Kahn New Rib Bars.—Same remarks as for corrugated bars. 

Havemeyer Bars.—Round and square Havemeyer bars have the 
same properties as the corresponding round and square plain bars. 
Since the projections form a part of the effective area, the weight of 
these bars is the same as that of plain bars. 

Havemeyer flats are of the dimensions given below. 


mize, in. | 1X 





‘AX S/d X 8/12 X geld X Bld X 2/18 X BLE X ILE X 9 


eI|-H 





























Area, sq. in.|0.2500!0.3750|0. 4690! 0.4688 |0. 5625/0. 7500/0.6563| 0.7656 [0.8750 
Weight per | : 
foot, UN es 1.280 





1.590 | 1.590 {1.913 |2.550 |2.230 | 2.600 |2.980 














14 REINFORCENEMT 


Elcannes Bars.—These are rolled in sizes from 2? in. to 14 in. 
with a variation of } in. The bars correspond in area and weight to 
square bars. 











Kahn Bars with Sheared Diagonals 


Fig. 2.—Sections of Kahn Trussed Bars. (See p. 13.) 


Herringbone Bar.—The sizes and weights of the Herringbone bars 
are as follows: 





Size, inches 13 14 1 z 3 5 





























Weight per foot, Ib...) 5.13 | 3.62 2.38 1.72 1.28 0.91 





Monotype Bars——These bars are rolled in one type only. The 
size, however, is arranged so that it is possible to get bars equivalent 
to standard sizes of round and square bars. The weight of bars is 
slightly higher than for plain bars because of the additional weight 
of lugs. 

Wire Used as Reinforcement.—The wire used for spirals in 
spiral columns, is cold-drawn from rods which are hot-rolled from 
billets. The cold drawing causes the wire to attain tensile strength 
from 80000 to 100000 Ib. per sq. in. It is usually designated by 














SPECIAL REINFORCEMENT 15 
gage numbers. The sizes of wire used in practice are given 
below. 

Gage | Equivalent | Area Gage Equivalent Area 
Number | Diameter, in.| Sq. in. || Number | Diameter, in. SG. ii: 
7-0 0.4900 0.189 5 0.2070 0. 034 
6-0 0.4615 0.167 6 0.1920 0.029 
5-0 0.4305 0.146 7 0.1770 0.025 
4-0 0.3938 0.122 8 0.1620 0.021 
3-0 0.3625 0.103 9 0.1483 0.017 
2-0 0.3310 0.086 | 10 0.1350 0.014 
0 0.3065 0.074" 4 11 0.1205 0.011 
1 0.2830 0.063 12 0.1055 0.009 
2 0.2625 0.054 13 0.0915 0.007 
3 0.2437 0.047 14 0. 0800 0.005 

4 O22535.,. 4 0.040 











SPECIAL REINFORCEMENT 


Special reinforcement consists of expanded metal and wire mesh 


of various sorts. 


Expanded Metal, shown in Fig. 3, p. 15, is produced by 
special machines from soft open-hearth steel plates. 
are slit and expanded by a cold drawing process in such a way that 


ao 


Fig. 3.—Expanded Metal. 


(See p. 15.) 


These plates 


in the final state the strands form diamond-shaped meshes. The 
cold drawing raises the elastic limit of the steel to around 55 000 lb. 


per sq. in. 


The standard sizes of expanded metal are given in the table 


on page 16. 


16 


REINFORCEMENT 


Standard Sizes of Expanded Metal Adaptable to Concrete Reinforcing 


Designation 


*3-13-075 
*3-13-10 
*3-13-125 





73-09-15 
*3-9-175 
*3-9-20 


*3—-9-25 
*3-9-30 
*3-9-35 


3-6-40 
3-6-45 
3-6-50 


3-6-55 
3-6-60 


*3-1-10 
*3-1-100 





Sectional Area, 
sq. in. per ft. 
Width 


0. 
0. 
0. 


0. 
0. 
0. 


0. 
0. 
0. 


0. 


0. 
0. 


Li; 


| 
75 


075 
10 
125 


15 
175 
20 


25 
30 
35 





Weight, 
Ib. per 
sq. it. 


0.27 
0.37 
0.46 


0.55 
0.64 
0.73 


0.92 
1.10 
1.28 


1.46 
1.65 
1.83 


2.01 
2.19 


2.74 
3.63 











Width of 
Standard 
Sheet 


6/ QO” 
6/ QO” 
5/ pues 


af! 0” 
6/ QO” 
5/ UN 


At Oe 
VE OQ” 
6/ 0” 


7 0!’ 
6’ Bet 
iy g”’ 


ye 3! 
Ae Q’’ 


iy OM 
4’ ae 





Number of 
Sheets in a 
Standard Bundle 


10 
7 
i 


5 
5 
5 


5 
2 
2 


2 
2 
2 


2 
2 


1 
1 


eS 


All of the above have a diamond-shaped opening, 3 by 8 in. 


All items listed above are furnished in 8, 12 and 16-ft. lengths. Items marked 
thus * are also furnished in 10-ft. lengths. 


* 3-14-75 and 3-1-100 are manufactured to order only. 


Fig. 4.—Welded Wire Fabric. 


‘ 
‘, 
Ary 


Welded Transverse Wires 
Ly 





No 
“; Longitudinal Wires 


(See p. 17.) 


SPECIAL REINFORCEMENT 


17 


Welded Wire Fabric consists of two sets of parallel cold- 
drawn wires placed at right angles to each other and electrically 
welded at the intersections. The wires are galvanized before welding. 


Welded Wire Fabric 















































| 
Transverse Sue ; Area in Longitudinal Wires Only, 
Wires | pe eainal Wy Ires per Foot of Width, sq. in. 
ae Brees: at Spacing of Longitudinal Wires 
Gage aes teed of wire, | of wire, |—— ) 
ae Seo inal aan es me ein: 
6 OK 0 0.307 0.074 | 0.443) 0.295) 0.221} 0.177; 0.148 
6 12 i 0.283 0.063 0.377) 0.252} 0.189} 0.151] 0.126 
8 Ly 2 0.263 0.054 | 0.325} 0.217) 0.162) 0.130) 0.108 
8 12 3 0.244 0.047 0.280; 0.187; 0.140} 0.112) 0.093 
9 12 4 0.225 0.040 | 0.239) 0.160) 0.120) 0.096) 0.080 
9 12 ‘i 0.207 0.034 | 0.202} 0.135) 0.101) 0.081} 0.067 
10 12 6 0.192 0.029 0.174] 0.116} 0.087} 0.069) 0.058 
igen cae, 12 7 0.177 0.025 0.148} 0.098} 0.074) 0.059) 0.049 
10 12 8 0.162 0.021 0.124, 0.082; 0.062} 0.049} 0.041 
12 Ml 12 9 0.148 0.017 0.1041 0.069} 0.052) 0.041} 0.025 
12 12 10 02135 0.014 0 0.043) 0.034) 0.029 


| 0.057 








Triangle Mesh.—Triangle Mesh woven wire reinforcement for 


concrete is made with either 
solid or stranded longitudinal 
members, properly spaced by 
means of diagonal cross wires, 
so arranged as to form a series 
of triangles between the longi- 
tudinal or tension members; 
the longitudinal members are 
spaced 4 in., the cross wires 
either 2 or 4 in. as desired, 
providing either a 2- or 4-in. 
mesh. ‘The sizes of both longi- 
tudinals and cross wires are 
varied in order to provide the 
cross-sectional areas of. steel 


required to meet the conditions. 





Fig. 5.—Triangle Mesh Wire Fabric. i 
(See p. 17.) 


18 REINFORCEMENT 


Information about triangle mesh may be taken from table below: 


Triangle-mesh Wire Fabric 


Number and Gage of Wires, Areas per Foot Width and Weights per 100 
Square Feet. 


Longitudinals Spaced 4 Inches. Cross Wires Number 14 Gage Spaced 4 Inches. 


ee ee ee EE St 
| Total Effective 











: Number and Gage ee tr ae Longitudinal | A ee Ni 
Style of Wires, each ace gitu oe S | Sectional Area Weight lbs. 
Number Longitudinal Square ches | square inches Ree 
per foot width per foot width 100 square feet 
032 1—No. 12 gage .026 .032 22 
.040 | ES .034 .040 20 
049 fo OS .043 .049 28 
058 Pe 3 ee .052 .058 32 
068 te ees .062 .068 30 
080 pene Uf tae 074 .080 40 
093 jo ee 087 .093 45 
107 be ra oe 101 .107 50 
126 1p ee Pal 20 oot 57 
146 Bt og .140 . 146 65 
153 1— + inch .147 . 153 68 
168 1—No. 2 gage . 162 .168 74 
180 Qt aioe 174° .180 78 
208 De ete ee 202 .208 89 
245 Dad. nek See 239 .245 103 
267 See Gan .261 . 267 111 
287 Se Og 281 287 119 
309 a tT ae 303 . 309 128 
336 Se aa aes 330 .336 138 
365 See a .359 .305 149 
395 pee Be 389 395 | 160 


Length of Rolls: 150-foot, 200-foot and 300-foot. 


Widths: Approximately 16-in., 20-in., 24-in., 28-in., 32-in., 36-in., 40-in., 44-in., 
48-in., 52-in. and 56-in. 

Norz.—Material may be furnished either plain or galvanized. Unless other- 
wise specified, shipments will be made of material not galvanized. 


SPECIAL REINFORCEMENT 19 


Expanded Metal Lath.—Expanded metal may be used as lath 
for plaster or stucco. . Two kinds of metal lath are obtainable, one 
painted black, and the other cut from galvanized sheet. The weights 
given in the table below are for painted lath. Galvanized lath 
weighs 0.4 lb. per sq. yd. more than painted lath. 


Expanded Metal Lath 

















‘ é pecan BO Size of pues Sq. Yd. in | Weight per 
Designation | sq. yd. in Sheet, in a Bundle Bundle, Ib 

Bundle, Ib. ’ "| Bundle ; Ge 
22-P 4.37 24 x 96 10 if cor detoaad aaa els 
24-F 3.40 24 x 96 15 26.66 90.67 
25K 3.00 24 x 96 15 26.66 80.00 
26-F 2.55 24 xX 96 15 26.66 68.00 
gpk 2.33 24 x 96 1b 26.66 62.22 
24-H 2.90 28 x 96 14 29.00 84.10 
26-H 2.20 28 x 96 14 29 00 63.80 

| | 





Rib Lath.—There are several types of expanded metal lath 
on the market which are provided with stiffening ribs running 
longitudinally with the lath. Rib lath is used for suspended 
ceilings, for partitions, 
and for other purposes. 
The function of the 
ribs is to stiffen the 
lath, so that the fur- 
ring strips or other sup- 
ports may be spaced 
farther apart. 

Figure 6, p. 19, 
shows two types of @S8RARRARAABRRAREESAARES 
Rib Lath, which are inch Hy oad Lath 
marketed under the 
trade name of Hy-rib. 
The ribs of the lath are 2 in. and ? in. deep respectively. The 
2-in. Rib Lath is used where greater rigidity and strength are 
required, 





Fic. 6.—Special Type of Rib Lath. (See p. 19.) 


CHAPTER III 
TESTS OF REINFORCED CONCRETE* 


The selected tests presented in this chapter were originally carried 
out to determine the principles of the theory and design of reinforced 
concrete. They are given here to illustrate the principles and con- 
clusions presented in the following chapters. 

The following data are obtained by tests: 

Stress at which first cracks appear and corresponding stretch in 
concrete; | 

Location of neutral axis; 

Relation between ultimate compressive fiber stress and strength 
of concrete in compression ; 

Distribution of stresses over a section; 

Relation between bending moment and moment of resistance 
based on stresses in steel; 

Effect of the percentage of reinforcement; 

Effect of mix of concrete and age. 

The results and conclusions are given on the following pages. 
Formulas for design based on the tests may be found in Chapter V, 
pp. 201 to 302. 

Behavior of Rectangular Beams During Loading.—When a 
reinforced concrete beam is subjected to a loading test its behavior 
is not the same throughout the progress of loading. Three stages can 
be distinguished: first stage, before the appearance of the first crack; 
second stage, after first crack is developed, but before either of the 
materials passes its elastic limit; third stage, after the elastic limit 
of steel or concrete has been passed. 


*To avoid breaking the continuity of treatment of theory and design this 
chapter on Tests is placed to precede the Theory instead of following it in regular 
sequence. When using the book as text book this Tests chapter may be taken 
up subsequently to the chapter on Theory preferably using it merely as supple- 
mentary material. Notation used in this chapter is explained in Chapter IV. 


20° 


BEHAVIOR OF RECTANGULAR BEAMS 21 


First Stage—Before the appearance of the first crack, a rein- 
forced concrete beam behaves like a homogeneous beam. Com- 
pression is resisted by concrete, and tension is resisted by concrete 
and steel in proportion to their moduli of elasticity. The position 
of the neutral axis nearly coincides with the center of gravity of a 
section obtained by replacing the steel by an area of concrete equal 
to the area of steel multiplied by the ratio of moduli of elasticity. 
This is called the first stage. It lasts until, at an elongation or 
stretch of concrete equal to the ultimate stretch of plain concrete, 
the first cracks appear in the beam. 

Second Stage.—This stage begins when the concrete cracks. 
At first, the cracks are not visible to the naked eye and do not extend 
up to the reinforcement. At increased loads, the number of cracks 
increases. They widen and move up toward the center of the beam. 
The neutral axis moves up. The larger the amount of reinforcement, 
the larger the number of cracks and the smaller their width, as illus- 
trated in Fig. 7, p. 23. In this stage, tension is resisted by steel 
and by the portion of concrete above the crack and below the neutral 
axis. 

The cracks never extend all the way up to the neutral axis (which 
rises as the cracks develop). The stretch is zero at the neutral axis 
and increases in a straight line to its maximum at the level of the steel. 
There must be, therefore, a portion of concrete below the axis where 
the deformation is smaller than the ultimate stretch for the concrete, 
and there must be a fiber which is just at the point of breaking. 
Above it, the concrete is intact and is carrying tensile stresses. 

The ratio of the tensile stresses carried by the concrete to the 
total amount of tension decreases with the increase in the load. For 
equal intensity of loading, the amount of stresses carried by concrete 
is larger for smaller percentages of steel. For large loads and per- 
centages of steel, commonly used in design, the amount of the stresses 
resisted by concrete is negligible. The assumption made in the 
design formulas, given in the chapters on Theory and Design, that 
steel resists all tension stresses and concrete in tension is disregarded, 
is, therefore, accurate enough for practical purposes. This assump- 
tion is still further justified by the fact that the tensile strength of 
concrete may be totally destroyed by shrinkage or temperature con- 
traction cracks and by construction joints. In analyzing tests, how- 
ever, it is necessary to take the tension in concrete into account, 
since its existence explains why the moment of resistance, based on the 


22 TESTS OF REINFORCED CONCRETE 

stress in steel and figured by formula M = A,fjd, is smaller than 
the bending moment, and why the actual stresses in steel obtained 
from deformations are smaller than the theoretical stresses. (See 
Fig. 9, p. 29.) The amount of tension carried by the concrete 
may be estimated by comparing the moment of resistance, based on 
the actual stresses in steel, with the bending moment. 

Third Stage. Beams Failing by Tension in Steel.—When the steel 
reaches its elastic limit, one or two of the cracks, which have been 
small up to this point, begin to widen and extend toward the top. 
This is shown by the loads underlined in Fig. 7, p. 23. The deflec- 
tion increases appreciably as the eracks widen and extend toward 
the top (the neutral axis rising), the compressive area becomes 
smaller, and finally the beam fails by the total destruction of the 
compressive area. This condition is brought about by a small 
addition to the load at which the steel passes the elastic limit. Thus, 
the passing of the elastic limit of steel marks the failure of the beam. 
Ultimate strength of steel 1s never reached, and is, therefore, of no 
consequence in reinforced concrete design. 

Beams Failing by Compression.—For beams failing by crushing 
of concrete, the third stage is marked by cracks in the top of the 
beam, which appear after the clastic limit of the concrete in com- 
pression has been reached. At increased load, wedge-shaped pieces 
of concrete spall off and the beam fails. 

Appearance of First Crack and Corresponding Stretch in Con- 
crete.—Numerous tests prove that the appearance of the first 
cracks in reinforced concrete corresponds to about the same stretch 
as in plain concrete. This stretch may be taken approximately as 
0.00012 of its length (corresponding. to 3 600 lb. per sq. in. in the 
steel) for 1: 2:4 stone concrete, and 0.00018 (corresponding to 
5400 lb. per sq. in. in the steel) for cinder concrete.’ At this 
stretch, however, as discussed below, the cracks are very minute 
and not visible to the naked eye. | 

The early conclusion of Mr. Considére in France, as the result of 
his tests, that the stretch of concrete when reinforced was 0.002 of its _ 
length, or about twenty times the stretch of concrete without rein- 

forcement, has been disproved by further experiments. Professor 
Turneaure,2 in testing moist beams, observed, at about the same 
stretch at which the first cracks developed in plain concrete beams, 


1 Technologic Paper No. 2, U.58. Bureau of Standards, 1912, p. 39. 
2 Proceedings, American Society for Testing Materials, 1904, p. 498. 


4 SJI0M INO T 


) 


a 
O 


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("Ge “a 29g) 
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ee \ ones 


ee 
PSE. xe 

Boos Ce 

- oe = 


seen & . gow 
Ci nace Ss oi 
ee 

# GOS! 


A 


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Y}IM syovsiy JO UOT}soOg UI UOIVUVA—) ‘DI 





24 TESTS OF REINFORCED CONCRETE 


dark marks which he called watermarks. Part of these watermarks 
developed later into actual cracks. Professor Bach * investigated 
the subject further and came to the conclusion that watermarks are 
places where adhesion between particles of concrete becomes weak- 
ened just previous to the formation of cracks. In plain concrete, 
each watermark develops into a crack. In reinforced concrete, on 
the other hand, only a part of the watermarks actually open, be- 
cause the steel strengthens these weakened spots and either retards 
the appearance of actual cracks or prevents their formation 
altogether. 

Professor Bach’s Tests.—Professor Bach’s tests ? in Stuttgart, 
summarized below, give the relation between actual and computed 
tensile stresses in concrete and steel just before the first crack, in 
beams with different mixes of concrete, different percentages of steel, 
and various conditions of storage. All of the values are high, as 
Bach evidently obtained a stronger concrete than the same propor- 
tions give ordinarily. It must be noted further, as has been empha- 
sized elsewhere, that the actual stresses in steel at this stage are | 
much smaller than the computed stresses. This does not affect the 
accuracy of the ordinary formulas for practical design, as it does 
not increase the factor of safety nor the load at elastic limit. 

The Influence of Mix of Concrete——The table on p. 25 gives the 
tensile stresses in concrete at first crack, the actual stresses in steel 


and the stresses computed by formula fs = iG As would be 


expected, the richer concretes resist larger stresses before cracking 
than the leaner concretes. The difference between computed and 
actual stresses in steel is also larger for rich concretes than for lean 
concrete. It is interesting, however, to note that the ratio between 
the computed stresses in steel and the actual stresses equals 35 for 
all mixes of concrete. 

Influence of Storage-—The tensile stress in concrete, fi, was 
smaller for beams stored dry than for beams kept wet, the difference 
amounting on an average to about 20 per cent. Concrete stored 
dry tends to shrink, causing initial tensile stresses in concrete because 
free movement of concrete is prevented by the adhesion of concrete 
to steel. Concrete kept wet, on the other hand, tends to expand, 
which, prevented by the steel, causes initial compressive stresses in 
concrete. When loaded, the initial tensile stresses increase tension 


3 Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Ausschuss, 
Heft 24, 1913. 


PROFESSOR BACH’S TESTS 25 


on the section while initial compressive stresses decrease it.. To 
concrete in building construction the values for dry storage are 
applicable because, even if the concrete is kept wet during construc- 
tion, in course of time it will dry out and the ultimate amount of 
shrinkage will be substantially the same as if it were held in dry 
storage. 


Actual and Computed Stresses at First Crack for Different Proportions of Concrete. 
(See p. 24) 


Age of beams at test, 45 days; aggregates, Rhine sand and gravel; ratio of 
steel, p = 0.0056. Wet storage. 


Compiled from tests by C. Bacu. 


Strength of Plain Concrete Tensile Stresses at First Crack 











Lb. per sq. in. Lb. per sq. in. 
In concrete In steel 
Proportions 

Compressive| Tensile Computed 
Actual by formula 

ft stresses M 
fs fs = het 

123:: 4 2 100 198 290 3 900 13 400 

REE be 3 750 270 380 5 150 17 200 

CR Gast 4 400 330 485 6 G00 23 000 


Tensile stresses, f;, and actual stresses, fs, are figured by formulas on page 207, 
in which the tensile stresses in concrete are taken into account. The stresses 


computed by formula, fs = , on the other hand, are figured neglecting the 


M 
Agia 
tensile value of concrete. 

Influence of Percentage of Steel.—Professor Bach’s tests* show 
that in concrete beams of the same proportions the actual unit stresses 
in concrete and steel at first crack are constant, irrespective of the per- 
centage of steel in the beam. ‘The theoretical stresses in steel at the 
first crack, however, figured by the ordinary formulas neglecting the 
tensile resistance of concrete, vary with the percentage of steel. 
Similar results, as shown in the table below, were obtained in the 
tests by the Bureau of Standards carried on by Mr. Richard L. 
Humphrey and Mr. Louis H. Losse.® 


4Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Aus- 
schuss, Heft 24, 1913. 
> Technologic Paper No. 2, U. 8. Bureau of Standards, p. 39, 


26 TESTS OF REINFORCED CONCRETE 


Actual and Computed Stresses with Different Percentages of Steel. (See p. 25). 


ee 


| 
Tensile Stresses in Steel at First Crack 


























Lb. per sq. in. 
Proportions 
Experimenters of = 
P Days|Actual}| Computed by ordinary formula 
Concrete 
stress 
lb. per 
sq.in.| p = .005 | p =.01 |p =.02 
Bachsuicinnncet eee 1522.5: 2) 45.1°6 600 |, 25 000 13 000 8 000 


Bureau of Standards.| 1:2: 4 28 | 4200} 25000 13 000 8 000 


Influence of Consistency.—A wet consistency reduces the strength. 
At an age of forty-five days for 1 : 2:3 concrete, p = 0.0056, and 
for percentages of water by weight, varying between 6.8 and 10.0 
per cent, the tensile stress, fi, at first crack ranged from 395 lb. to 
310 lb. per sq. in., while the compressive strength of the same con- 
crete ranged from 3800 Ib. to 2360 lb. per sq. in., and the tensile 
strength in direct pull, from 485 Ib. to 245 Ib. per sq. in. 

Influence of Age.—Increase in strength with age, as determined 
by Bach, is shown in the table below. 








Actual and Computed Stresses at Different Ages. (See p. 26) 
Proportions of concrete, 1 : 2:3. Ratio of steel, p = 0.0056 
Compiled from tests by C. Bacu * 


an a a ee er va 
| 


Stresses at First Crack 
Lb. per sq. in. 








Age Age Age Age 








28 days 1 45 days | 6 months | 1 year 





f, Actual stresses in concrete... .| 360 380 466 495 


f, Actual stresses in steel....... | 4900 5 100 6 300 6 700 
fs Computed stresses in steel....} 16 400 17 500 21 700 23 000 


ee ee eee 


The computed stresses, fs, are the stresses in steel figured by Formula (7), p. 
207. 


* Bach. Spannungen unmittelbar vor der Rissbildung. Deutscher Ausschuss, Heft 24, 1913. 


POSITION OF NEUTRAL AXIS 27 


Position of Neutral Axis.—The position of the neutral axis in rein- 
forced concrete beams varies with the percentage of steel and the 
strength of concrete and also with the intensity of the loading. For 
beams with large percentages of steel, the initial position of the neutral 
axis is lower than for beams with smaller percentages of steel. With 
the same steel and stronger concrete, the neutral axis is higher than 
with a weaker concrete. 

In any beam, the neutral axis at the beginning of the loading 
nearly coincides with the center of gravity of a section in which the 
steel is considered as replaced by an area of concrete equal to the 
area of steel times the ratio of the moduli of elasticity. With the 
progress of the loading, it moves upward. Figure 8. p. 27, illus- 


arn 





oe 
Moment | Computed = f 
‘A : tae 7 ' y 20 
Initial = Moment S44 center Line ' Computed ~ 














= 
= al tA = 
Center Line ; H | Moment.7y44 Center Line 
4 Initial lean O75) ae 
i] ' 
“4 ‘ ' Initial 
S ' ' Crack, 
& : ‘ f 
3 “gon i gt ' " 
ye Bar | oy Bar: | 1 Bar 
- ———— - - 


Fic. 8.—Change in Position of Neutral Axis During Loading for Different 
Percentages of Steel.° (See p. 27.) 


trates the typical movement of the neutral axis during loading for 
beams with different percentages of steel. As is evident from the 
figure, the position of the neutral axis for different loadings was deter- 
mined by plotting at proper levels the deformation of the upper 
concrete fiber and the deformation of steel. The intersection of the 
line obtained by connecting the two points and the vertical section 
of the beam gives the position of the neutral axis. For usual per- 
centages of steel, the distance from the compressive side of the beam 
under working loads is three-tenths to four-tenths of the depth. 
The formula for location of neutral axis is given on p. 129. 

Stresses in Steel for Varying Intensity of Load.—Figure 9, 
p. 29, gives the typical deformation of steel and of the upper fiber of 
concrete in inches per inch of length in beams with different per- 
centages of steel, based on the tests of Messrs. Humphrey and Losse.’ 


6 Bach ‘“‘Biegeversuche mit Eisenbetonbalken,” Berlin, 1907, pages 7 and 8. 
7 Technologic Paper No. 2, U. 8. Bureau of Standards, 1912, p. 39. 


28 TESTS OF REINFORCED CONCRETE 


The deformation curve for steel is not a straight line, but a com- 
posite curve, the shape of which varies with the percentage of rein- 
forcement. The deformation and, therefore, the stresses in steel at 
the first stage of the loading, that is, before the first crack, are com- 
paratively small and proportional to the load, so that the deformation 
curve for this stage is almost a straight line. At deformation equal 
to the ultimate deformation in plain concrete beams, cracks appear 
in the concrete, and the tensile stresses borne by it are transferred 
to the steel, causing an abrupt change in the steel deformation curve. 
As is evident from the change in deformation, as shown in the dia- 
gram, the change in the direction of the curve is much larger for 
smaller percentages of steel, because the amount of tensile stress, 
constant for beams of equal cross sections, which is transferred from 
the concrete to the steel, is distributed over a smaller amount of 
steel, and the increase in the unit stress in steel is therefore larger. 

On the deformation diagram, the load at first crack is marked by 
the change in deformation from a straight line to a curve. 

After the first crack, a large proportion of the total tensile stresses 
is carried by the steel. The concrete, however, still carries a small 
proportion of tensile stresses, dependent in amount upon the per- 
centage of reinforcement in the beam. Because of these stresses 
carried by the concrete, the deformation in steel at different intensi- 
ties of loading does not vary proportionally to the load. | 

It is absolutely necessary that this be taken into account in 
analyzing results from tests not carried to the breaking point, for 
instance, in tests of completed buildings. 

The actual stresses in steel corresponding to the deformation are 
smaller, for reasons indicated above, than the stresses computed by 
formula f; = SEN for the same load. In beams with small per- 
centages of steel, concrete carries a considerable portion of the 
stresses up to the breaking point of the beam. (This is shown by 
the deformation curve for beam with 0.49 per cent of reinforcement.) 
For larger percentages of steel, the dash line on the diagrams, which 
indicates the theoretical deformation of the steel obtained from 
Formula (9), p. 131, strikes the actual deformation curve at the 
deformation corresponding to a stress of 39 000 Ib. to 48 000 Ib. per 
sq. in. This indicates that, near the elastic limit, the actual stresses 
agree very well with the theoretical stresses. 


The fact that at design load the actual stresses in steel are much 


STRESSES IN STEEL FOR VARYING INTENSITY OF LOAD 29 


lower than those obtained from formulas may create a wrong impres- 
sion as to the actual capacity of the beam. To get the right idea, 


ie 

S 
eee ae : oe : : : : ; : 
Deformation Per Unit of Length Deformatien Per Unit of Length 
Beam With 0,49 PerCent Reinforcement Beam With 0,98 PerCent Reinforcement 


4 











lal 


hy 
V/ 
































3 
I~} 
é 
a 
8 
os} 
8 
= asl 
edhe ies c 
Peerarey (key 
bsliea te co ek meal 
aig 
ha bait ie Bilis 
ia ic bash 
Weight -+ isl had afta lies 
of Beam © 
Defatmation Per Unit of Length a3 Detbrmek Per Unit ire 
Beam With 1.30 PerCent Reinforcement Beam With 1.72 Per€ent Reinforcement 


Fic. 9.—Deformations in Steel and Concrete Due to Loading.* (See p. 27.) 


it must be remembered that the conditions at the elastic limit, and 
not those at the intermediate loads, govern the strength. It is 
* Technologic Paper No. 2, U. S. Bureau of Standards, 1912, p. 39. 


30 TESTS OF REINFORCED CONCRETE 


desirable to get a certain factor of safety. This means that it is 
required to build the beam so that it will not fail under a smaller 
load than the design load multiplied by the factor of safety. When 
the distribution of stresses in a beam varies, i.e., when the distribu- 
tion at the design load is different from the distribution at the elastic 
limit, two methods of procedure are possible. The first method is to 
multiply the design load by the factor of safety and then use the dis- 
tribution and magnitude of stresses at the elastic limit. The second 
method is to use the design load in computations with unit stresses 
equal to the elastic limit divided by the factor of safety. However, 
the distribution of stresses used in this method is the same as exists 
at the elastic limit. ‘The second method is commonly used, with 
the understanding that the stresses used in design indicate only 
the factor of safety and not the actual stress. 

The formulas on p. 131, therefore, although they do not repre- 
sent the actual conditions of stresses at the design load, give the 
required factor of safety and are recommended for use in design. 

The behavior of a beam reinforced with steel in two layers is 
substantially the same as that of a beam with one layer. The only 
difference is that the steel is not as effective as if all the bars were 
placed in one layer. The bars in the second layer are nearer the 
neutral axis. Their moment arm is, therefore, smaller than that of 
the bars in the first layer. Also, their unit stress is smaller because 
it varies with the distance from the neutral axis. 

The method used in design where the depth of the beam is taken 
as the distance from the center of gravity of the bars to the top, is 
not exact, because it does not take into account the difference in 
stresses between the two layers. It is, however, exact enough for 
practical purposes. 

Tests of Beams Failing by Compression.—From tests of beams 
failing by compression, the following conclusions may be drawn: 

The computed compression stresses in the extreme fiber in con- 
crete, at the load causing failure, are much larger than the strength 
of concrete in compression obtained from control cylinders of stand- 
ard dimensions (8 X 16 in.) made of the same material as the beams, 
cured in the same way, and tested at the same time. As would be 
expected, the difference is much greater for the stresses obtained 
from formulas based on straight-line distribution of stresses than 
from formulas based on parabolic distribution. 

The difference between the fiber at failure and the strength of 


TESTS OF BEAMS FAILING BY COMPRESSION dl 


control cylinders may be explained as follows: In a beam, only the 
extreme fibers are exposed to the maximum stresses. The adjoining 
fibers, being nearer the neutral axis, are subjected to smaller stresses. 
If any weakness should develop in the extreme fibers in any part of 
the beam, causing them to reach the point of crushing, actual crush- 
ing would not take place, because the less heavily stressed adjoining 
fibers would come to the assistance of the heavily stressed fibers and 
would assume a part of the excess stress without exceeding their 
ultimate strength. <A series of such adjustments of stresses is likely 
to take place before final failure. 

In a cylinder, on the other hand, all the concrete is subjected to 
the same stresses at the same time, so that weak particles in a cyl- 
inder cannot be assisted by the adjoining particles without i increasing 
their unit stresses. The compressive stresses are ‘not evenly dis- 
tributed over the whole area because the material is not perfectly 
homogeneous. Deformation readings on opposite sides of a cylinder 
often show that the compression on one side is different from the 
compression on the opposite side. This uneven distribution of 
stresses not only increases the stresses of some particles of the cylin- 
der over the average stresses, but also produces harmful shearing 
stresses in the cylinder, which often cause final failure. 

From the above description of the conditions in a beam and 
cylinder at failure, it follows that some difference between the 
actual fiber stresses at failure and actual unit strength of cylinders 
is to be expected. The difference is not as large, however, as it is 
shown to be by computations. The large difference between max- 
imum compression strength in the cylinder and in a beam is largely 
due to the inexactness of formulas for computing fiber stresses. The 
stresses depend largely upon the magnitude of the modulus of elas- 
ticity, which varies not only for different concretes, but also for dif- 
ferent intensities of stresses in the same material. The straight- 
line formula especially, being based upon the assumption of constant 
modulus of elasticity, gives much larger stresses than are actually 
developed in a beam. This is compensated by accepting larger unit 
stresses with straight-line formulas than would be permitted for formu- 
las with varying modulus of elasticity, such as the parabolic formulas. 

Details of tests of beams failing by compression are given in a 
paper by Messrs. W. A. Slater and R. R. Zipprode’ The tables 
on pp. 32 to 34 are taken from this paper. 


®’ Compressive strength of concrete in flexure by W. A. Slater and R. R. 
Zipprode: Proceedings, Am. Con. Ins., 1920. Vol. XVI, p. 120. 


TESTS OF REINFORCED CONCRETE 


o2 


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66100'0 | ¢9000'0 
6£Z00'0 | €6000'0 
€8100'0 | €2000°0 
Zcc00°0 | 28000°0 
10Z00°0 | 28000 0 
Z8100°0 | 89000°0 


9€200°0 | 89000°0 
¢9Z00°0 | 69000°0 
#6100'0 | c7000'0 
€1Z00'0 | 6F000'0 
89200°0 | 89000°0 
#LZ100°0 | 92000°0 
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TESTS OF T-BEAMS 35 


TESTS OF T-BEAMS 


The discussion of the phenomena of loading and the movement of 
the neutral axis, given for rectangular beams, and the results given in 
connection with the appearance of first cracks, on p. 22, apply also 
to T-beams. ‘The initial position of the neutral axis, however, will 
not be the same in a T-beam as in a rectangular beam and will depend 
upon the relative dimensions of the flange and the stem and the per- 
centage of reinforcement. 

It must be remembered, in applying to T-beams the discussion of 
influence of percentage of steel upon the appearance of first crack 
given for rectangular beams, that the percentage of steel must be 
figured for the width of beam equal to the width of the stem. 

T-beams may fail by tension in steel, compression in concrete, 
diagonal tension, or bond. The tests discussed below are grouped 
according to the cause of failure. 

Tensile Failures of T-Beams.—Professor Talbot’s test of T-beams ? 
consisted of nine beams. Dimensions: total length, 11 ft.; test 
span, 10 ft.; depth to steel, d = 10 in.; height, A = 12in.; thickness 
of slab, ¢ = 37 in.; breadth of stem, b’ = 8 in.; width of flange, 

= 16, 24, and 32 in. (three beams of each width). Concrete, 
1:2:4 by volume. Steel:. the amount of reinforcement varied 
from 0.92 per cent to 1.1 per cent of the area of enclosing rectangle, 
bd. Longitudinal reinforcement: ?-inch plain round bars with yield 
point of 38 300 Ib. per sq. in., and 3-inch corrugated square bars with 
yield point of 53 800 Ib. per sq. in., with } in. U-shaped stirrups 
(corrugated square) spaced 6 in. apart in the outside thirds of beam. 
All beams failed by tension. Stresses in steel at maximum load, 
M 
0.864.a’ Bere well with 
stresses at yield point of the steel. Calculated stresses ranged, for 
plain bars, from 37 600 to 41 500 Ib. per sq. in., with an average of 
39 800 lb., and for corrugated bars, from 55 700 Ib. to 64 300 Ib., with 
an average of 55 700 Ib. per sq. in. 
No beam failed by diagonal tension, although the maximum 


figured by Professor Talbot by formula, f, = 


17 
shearing unit stress from formula, v = aia’ reached the value of 


605 lb. per sq. in. The web reinforcement, therefore, proved to be 
adequate. The total diagonal tension, considered as resisted by the 


* University of Illinois Bulletin No. 12, February 1, 1907. 


36 TESTS OF REINFORCED CONCRETE 


stirrups only, would produce a theoretical stress in stirrups oi 
55 500 lb. per sq. in., or higher than the elastic limit of stirrup steel. 
Judging from the size of the diagonal cracks, the actual stress in 
stirrups was much below the elastic limit, which indicates that a 
part of the diagonal tension is carried by concrete, justifying the 
recommendation on p. 246 allowing part of the total diagonal ten- 
sion to be considered as resisted by concrete with the remainder 
carried by the steel. | 


Tests of T-Beams to Determine the Effective Width of Flange.— 
The following test was made at the testing laboratory in Stuttgart,!° 
with a number of beams of the same span, cross section, and amount 
of steel, but varying widths of flange. Three beams of each type 
were tested. Loads were applied at one-third points. 


T-Beam Tests to Determine Effect of Width of Flange. (See p. 36.) 
Compiled from tests by C. Bacx *° 


Proportions of concrete, 1: 3:4 by volume, with 93 per cent of water by 
weight. Elastic limit of steel, 48 000 lb. per sq. in. Age of test, 45 days. 

Common dimensions: Total length, 10.89 ft.; testing span, 9.84 ft.; breadth of 
stem b’ = 7.08 in.; total depth, h = 9.84 in.; depth of steel, d = 8.66 in.; thick- 
ness of flange, ¢ = 2.36 sq. in.; 2.36 in. fillets at juncture of slab and stem. 

Steel: 4-1.17 in. round bars; area of steel A, = 4.38 sq. in.; i-in. round U 
stirrups spaced 3 in. on centers in outside thirds. 

Beams 3, 4, and 5 were provided in outside thirds of beam with }-in. round 
cross bars spaced 6 inches on centers. Beam 5a had no cross bars. 


ee a a eee 























Ratio of Stresses at Maximum St renitai Ieee ee 
Width) Projections} Maxi- Load 6 : 
No. ; of Compressive 
of | of Flange | mum Lb. per sq. in. ; ; 
of Cubes | Fiber Stress 
Flange to Load 
Beam| . ‘ 7 ee to Strength 
in. | Thickness Lb. per of Cubes t 
of Slab Lb. ee fos sq. in. 
2 ol 0.0 16900} 10050 2 200 1 580 1.39 
a 4 201,90 5.0 31500} 18 700 1 900 1 580 1.20 
Ae 2021 9.5 47200} 28 100 2 200 1 750 1.26 
5 | 39.4 13.6 56 800} 33 750 2 140 1 800 1.19 
5a | 39.4 13.6 35 300} 21 000 1 340 1 620 0.83 


* Based on Formulas (9) and (11), p. 131. 
_t Ratios would have been still larger if oblong cylinders had been tested instead of cubes. 


0. Bach. Mitteilungen tber Forschungsarbeiten aus dem Gebiete des 
Ingenieurwesens, Hefte 90 and 91. 


TESTS OF T-BEAMS od 


Beams 2, 3, and 4 failed by crushing of concrete in the flange. 
The failure in concrete occurred as in cubes, by splitting of wedge- 
shaped pieces of concrete. The shortening of the flanges of Beam 3 
was uniform throughout the width of the flange during the whole 
progress of the test. For Beam 4, there was a difference in shorten- 
ing of only 8 per cent at loads near the crushing strength of concrete. 

Beams 5 and 5a were of the same design except that cross bars, 
spaced about 6 in. apart, were used in Beam 5 while 5a had no cross 
bars. Beams of both designs failed by the shearing off of the flanges; 
but there was great difference in action and in maximum load between 
the two beams, which is attributable entirely to the effect of the 
cross bars. In Beam 5, with cross bars, the stresses and the short- 
ening of the flange were uniform over the whole width of the flange 
up to a load of about 44 000 lb., at which point the first longitudinal 
' crack appeared at the junction of the slab and the stem. This crack 
was followed by a number of similar cracks, which finally caused 
failure at-a load of 56 800 lb. After the first longitudinal crack, the 
shortening at the edges, as compared with that at the stem, decreased. 
At the maximum load, the shortening at the edges amounted only 
to half of the shortening at the stem. Both edges of the flange of 
the T-beam bent down along the entire length of beam. The max- 
imum downward movement of the edge was in the center of the beam, 
where the edge of the flange was 0.012 in. below the level at the stem. 

Beam 5a, without cross bars, failed from the same causes as Beam 
5, except that the first crack occurred at a smaller load, and failure, 
caused by the separation of the stem and the flanges, followed closely 
the appearance of the first crack. This proved conclusively that it is 
advisable to place reinforcement across every beam that is expected 
to act as a T-beam s0 as to insure it against the separation of the 
flange and the stem. 

The compressive stresses in the flange, which must be transferred 
to the stem, cause shearing stresses along the juncture of the stem 
and the flange. The magnitude of the unit shearing stress is pro- 
portional to the amount of compression carried by the flange and 
inversely proportional to the thickness of the flange. For method 
of determining the shearing stresses, see p. 143. 

The following conclusions can be drawn from the tests: 


(a) Maximum load is increased materially by introduction of 
cross bars. 


38 TESTS OF REINFORCED CONCRETE 


(b) Maximum load can be still further increased by fillets. 
Beams with fillets, making a 30° angle with the hori- 
zontal and having a depth of three-fourths of the depth 
of slab, withstood a 20 per cent larger load than beams 
without fillets. The effectiveness of the fillet did not 
increase with the increase in size of the fillet. 

(c) No appreciable difference is shown in deformation in 
flange at the edge and at the stem for 39.4-in. flange 
(projections 13.6 times depth of slab). 

(d) Maximum load for beams with 60-in. flange (projections 
twenty-one times depth of slab) is only 5 per cent larger 
than for 39.4-in. flange (projections 13.6 times depth of 
slab). 


From the above conclusions, it is evident that: 

(1) In computing effective strength, the use of a width of flange 
equal to six times the thickness of the slab on each side of 
the stem is conservative. 

(2) In ordinary cases, no fillets are required. 

(3) Crossbars on the top of T-beams are required to insure T- 
beam action. 


TESTS OF REINFORCED CONCRETE BEAMS TO DETERMINE EFFECT OF 
DIAGONAL TENSION 


Very comprehensive diagonal tension tests, illustrated and 
described below, were made by Professor Bach in 1908 and 1912." 
They comprise sixty-four sets of beams divided into two groups. In 
Group I (see Fig. 10, p. 41), the load is applied to the one-third 
points, and in Group II (see Fig. 11, p. 43), at eight points, uni- 
formly spaced. Both groups include beams with no web reinforce- 
ment, with stirrups, with bent bars, and with stirrups and bent bars. 
The span of beams in Group I is 9.8 ft., and in Group IL 13.1 ft. 
Other dimensions are (using notation on page 215) h = 15.7 in.; 
d = 13.9 in.; b’ = 7.9 in. (except as noted); ¢ = 3.9 in. for both 
groups. The width of flange is b = 19.7 in. in Group I, and 6 = 23.6 
in. in Group II. The amount of longitudinal steel is practically the 
same in all the beams (see summary), the only variable being the 
arrangement of web reinforcement. 


11 Deutscher Ausschuss fiir Hisenbeton, Hefte X, XII, XX. 


TESTS OF REINFORCED CONCRETE BEAMS 39 


Concrete.—1 :2:3 by volume; 9 per cent of water by weight. 
Aggregates: Rhine sand and gravel. Age of specimens, forty-five 
days. Average strength of cubes, 3 440 Ib. per sq. in. Steel yield 
point varied from 45 400 Ib. to 51 000 Ib. per sq. in. for Group I, and 
from 44 000 Ib. to 63 000 Ib. per sq. in. for Group II. 

In both groups there are: 


(a) Beams without stirrups, with widths of stem, b’, 5.9 in., 
7.9 in., and 11.8 in. 

(b) Beams with U-shaped 0.275-in. round stirrups, spaced 3.8 
in. apart, and with widths of stem varying as in the 
previous case. 

(c) Beams without stirrups, horizontal steel provided with 
hooks at ends. , 

(d) Beams with U-shaped stirrups varying in diameter from 
0.2 in. to 0.39 in. and spacing varying from 2 in. to 
7.01, 

(e) Beams with bent bars of different arrangement, with and 
without sitrrups. 


In Group II, the web reinforcement was designed by Formula 
(66), p. 155, for loads producing in tensile steel a stress of 14 200 Ib. 
per sq. in. 

The figures on pp. 41 and 43 give the amount and arrangement 
of reinforcement, for selected beams from Groups I and II, respec- 
tively. The tables on pp. 42 and 44 give maximum loads carried 
by the beam; cross section of horizontal steel; cross section of 
bent-up bars; maximum shearing unit stress, v; stress in steel at 
maximum load, fs; stress in stirrups figured with assumption that 
stirrups take the total amount of diagonal tension. 

The following conclusions can be drawn: 

General.—(1) Tests show conclusively that it is possible to pro- 
vide sufficient web reinforcement, in the shape of stirrups, bent bars, 
or a combination of the two, to develop the maximum carrying 
capacity of the beam, whether governed by horizontal steel or by 
_ crushing strength of concrete. | 

(2) For beams with and without stirrups, failing by diagonal 
tension, the maximum load increases in direct ratio with the width 
of the stem (Group I (a) and (b) ). With same arrangement of stir- 
rups, the wider beams failed by diagonal tension at higher loads 
than the narrower beams. This proves that diagonal tension is 


40 TESTS OF REINFORCED CONCRETE 


resisted partly by concrete and partly by reinforcement; otherwise, 
in beams with stirrups failing by diagonal tension, the width of stem 
would be of no influence on the maximum load. 

(3) The stresses in web reinforcement are smaller than those 
obtained by assuming the total diagnoal tension from Formula (62), 
p. 149, to be resisted by steel only. T he two above conclusions justify 
the recommendation that, in designing, part of the diagonal ten- 
sion be considered as resisted by concrete. (See also p. 36.) 

(4) Hooks at ends of horizontal bars increase the strength of 
beams by preventing slipping of bars. (See also p. 60). The 
increase is dependent upon the ratio of the horizontal bars to total 
reinforcement. In case of Beams 1 and 7 (Group I) and Beams 51 
and 54 (Group II), where all bars were either straight or hooked 
at ends, the increase caused by hooking amounted to from 40 to 50 
per cent of the load carried without hooks. In other cases, where 
only a small amount of steel was affected, the increase varied from 
5 to 22 per cent. 

Stirrups—(5) Stirrups increase the capacity of the beam, as is 
evident from comparison of Beam 7 and Beam 16, on p. 42. For 
equal spacing, the ultimate load increases with the increase in the 
diameter of the stirrup; and for equal diameters of stirrups, it 
increases with the decrease in space between stirrups. 

(6) Stirrups of small diameter, spaced closely, are more effective 
than those of large diameter with correspondingly larger spacing. 
Stirrups in tests were most effective with a spacing equal to one- 
third of the depth of the beam; ie., for this spacing, the increase 
in ultimate load per pound of steel in stirrups was a maximum. 

(7) Only about one-third of the total depth may be counted 
upon in developing bond in stirrups, because with the progress of 
the diagonal cracks, which may reach to about one-third of the 
depth of the beam from the top, only the portion of concrete above 
the crack is effective. Slip, in stirrups under load, was largest, by 
actual measurement, for the stirrups intersected by the crack near 
the top of the beam. It is advisable, therefore, to hook the ends of 
the stirrups. 

(8) The stirrups influence the bond of the horizontal steel. The 
bond increases with the increase in the number of stirrups. 

Bent Bars—(9) Bars bent at one point only are more effective — 
when bent at about 45° than if bent flatter at about 18° with the 
horizontal. Beam 29, on p. 42, resisted 20 per cent larger load than 


TESTS OF REINFORCED CONCRETE BEAMS 4] 


Position of Loads 







ony a " Round Bars.” 





9 7) 4 
2-17— Round Bars. 


2-138" Round Bars.” 





: a Round Bars--“-7 


¢ 


| - L1"Round Bar 92. 15 Round Bars 





qe ie "Round , Bar> “2- ee Round Bars “1-1 x Round Bar 
yl-l z" Round Bar PEE ” Round Bar 45° 


2. a Round Bars --~ 


ver a Round Bar 





-3 » ie "Round Bars 3 -4 "Round Bars’ 


Fig. 10.—Effect of Diagonal Tension. Design and Loading of Test Beams. 
Selected Beams from Group I. (See p. 38.) 


42 


TESTS OF REINFORCED CONCRETE 


Beam 25, although the area of bars bent at 45° was 1.8 sq. in. against 
1.98 sq. in. bent at 18° in Beam 25. No marked difference was found 
in strength for beams with bars bent at 30°, 40°, and 45° respectively. 


Tests of Beams to Determine the Efficiency of Web Reinforcement. 


p. 38.) 


Compiled from Tests by C. Bacu * 


a On Sn enn 














Area of Cross 
: Stresses at 
Section of 
B : Maximum Increase in 
ars j , 
Load Maximum 
, Load by 
3 , Maxi- Ib. per sq. in. |Hooking of 
® | Hori- ; > 
faa! mum Horizontal 
., |zontal| Bent Bo 
(total) | 
ta 
& i iF | Ts v 
>, sq. in. |Sq. in. | | . Per cent 





nn ee 


Beams Loaded at Two Points 
| 











(See 
Deflection 
Cause 
of 
XI- eae Failure 
maxi- 
UM) Paes (See 
load eal note) 
in. | in. 
0.09 | 0.03 D 
0.19 | 0.06 D 
0°35) O41 D- 
0.28 | 0.09 | : D 
0.45 | 0.14 D 
0.38. 40218 D 
0.47 '|. 0.15 T 
0.41 | 0.14 D 
0.42 | 0.15 D 
0.47 | 0.16 ft, 
0.41 | 0.14 D 
0:36 4-081 D 


11°38: OLS eee 35 900} 990) 14630) 185 | ........ 
TO. G1 Baa 54 300} 1 580} 22 580) 286 iy 

164 Sr9l Bee 88 000] 2 430] 36 1380) 451 | ........ 
25 | 3.83 | 1.92] 75 900) 2 210} 32 980) 407 22 
27 | 3.83 | 1.92] 98 600] 2 860) 42 760) 528 | 20 
29 | 3.91 | 1.78} 92 400) 2 680) 39 500) 501 19 
30 | 3.91 | 1.78] 106 900) 3 090) 45 480 567 | ......-- 
32 | 3.91 | 3.57} 98 600) 2 940} 42 350) 529 | ........ 
36 | 3.91 | 2.64] 101 200} 2 940) 42 960} 535 12 
38 | 3.91 | 2.73] 108 900] 3 160| 45 830) 573 11 
AQ | 3.95 | 2.71) 100 100} 2 880) 41 690) 529 5 
50 | 3.87 | 1.77| 81800] 2 440) 35 340 488 | ........ 
Sor anes TI one nv 

Nore: D = diagonal tension failure 


B = bond failure 
bond failure 
tension failure 


es) 
I 


Jia 


* Deutscher Auschuss fiir Eisenbeton, Hefte X, XII, XX, 1908 and 1912. 


+ Areas of bars are converted directly from the metric dimensions. 


10 are approximate to nearest sixteenth inch. 


Diameters in Fig. 


TESTS OF REINFORCED CONCRETE BEAMS 43 


Position of Loads 











i) 


‘ag -] 2" Round Bars 





52 








\Q- 12° Round Bars 


53 





\.9-1;2" Round Bars 














2-1" Round Bars ~~--. 9.11" oy" aed 
a5 Round Bars aay Round 45° 45 


60 








‘ 3 4 : nn u“ ” 
-2-7"-7- 14 54 sr y 8 Round Bars tt Round. 
















W, é ‘ any yas. 
DD hy gy 19" Sy TI" Ny 9 
8 Eat 16 a 16 






" “ 
Round Bars -t Round 







-4 Ny SSM! 4 Sg", ey Ny -2"Round Bars 30° 35 et 745° 


, \ \ 











Fig. 11.—Effect. of Diagonal Tension. Design and Loading of Test Beam, 
Selected Beams from Group II. (See p. 38.) 


44 TESTS OF REINFORCED CONCRETE 

(10) Bent bars, as well as stirrups, are effective reinforcement for 
diagonal tension. Compare Beams 30 and 38, p. 42, both of which 
failed by tension in steel. 

(11) The strength of beams with bars having sharp bends was 
smaller than for beams with a circular bend with a radius equal to 
about 12 diameters. 

(12) It is evident from comparison of the stresses at maximum 
loads with the elastic limit of steel that almost all the beams with bent 
bars failed by tension in longitudinal steel. 





























Tests of Beams to Determine the Efficiency of Web Reinforcement. (See p. 38.) 
Compiled from Tests by C. Bacu * 
Area of (roast : 
Section of Stresses a Deflection 
Bars t Maximum Increase at 
Load Maximum 
Maxi- Load by Cause 
5 one | mum lb. per sq. in. Hooking of Mat Half of 
aia Load Horizontal maxi- |Failure 
., |zontaliBent mum 
S | (total) Bars eee mum | (See 
g Ib. load | note) 
S| te ae fe fs v eae 
3 ag in. |sq. in. Per cent in. in. 
! : : l ! 

Beams Loaded at Eight Points 
nL nner 
Ble SOs ake 46 900} 1°31.0)'22 420):252 | 222 eee 0.22 | 0.10 | DB 
2) sooo) ae ee 67 300} 1 870} 32 020) 360 | ...2.505 0.37.1 0.15 |; DB 
S| 33 a0) ol eee 51 300] 1 420} 24 390} 273 | .......: 0.24 | 0.09 D 
G4. yoeO7 Ree eee 93 900} 2 552| 43 250} 492 | ........ 0.73 | 0.22 
55 | 3.81 | 1.91] 73 300] 2 150) 36 890) 404 | ........ 0.50 | 0.20 8) 
56 | 3.80 | 1.91} 100 300] 2 920] 50 300) 549 | ........ 1.01 | 0.27 ik 
58 | 3.86 | 2.32| 95 300) 2 750) 46 690; 516 6 0.74 | 0.25 ie 
60 | 3.91 | 2.37] 95 300) 2 760) 46 220) 518 4d 0.64 | 0.24 TL 
62 | 3.94 | 2.74} 99 400) 2 860] 48 010; 539 17 0.88 | 0.26 - 
64 | 3.89 | 1.70} 106 200) 2 950} 50 450; 566 18 0.88 | 0.26 T 
66 | 3.91 | 2.73) 101 900) 2 900] 49 180) 552 | ........ 0.75 | 0.26 T 

| 
ee eee 
Note: D = diagonal tension failure 
B= bond failure 
T = tension failure 


DB = diagonal tension and bond failure 


* Deutscher Ausschuss fiir Eisenbeton, Hefte X, XII, XX, 1908 and 1912. 
+ Areas of bars are converted directly from the metric dimension. Diametersin Fig. 11 
are approximate to nearest sixteenth inch, 


45 


TESTS OF REINFORCED CONCRETE BEAMS 


(‘OF 


d aay) 


HOvg ‘doug Aq sysoJ, 


HOVG ‘doug Aq 84sa J, 
oN’ | UOTSUAT, [BUOSIC, [V 


(op ‘d aay) + ‘omnpreg uorsuay, TeoIdA,— eT “D1 


dA L— ZI. 





DI] 





46 TESTS OF REINFORCED CONCRETE 


BEHAVIOR OF REINFORCED CONCRETE BEAM FAILING BY DIAGONAL 
TENSION UNDER LOAD 


The difference between the intensity of loading at first diagonal 
crack and the ultimate loading, for beams without web reinforce- 
ment, depends upon the strength of the concrete. Lean, or green, 
concrete beams fail with little or no warning, so that the load at 
first diagonal crack coincides with the breaking load; whereas in 
richer and stronger concrete beams, diagonal cracks are visible for 
some time before final failure occurs. 

Figure 12, p. 45, shows a photograph of a beam of 1:2 :3 
concrete, forty-five days old, with no stirrups, after failure by diag- 
onal tension and slipping of the bar. The beam was loaded at one- 
third points. At a load of 14700 lb., the first crack developed in 
the middle portion of the beam. At 17 640 lb., a diagonal tension 
crack developed in the outside third, just beyond the load. ‘This 
crack increased with the load and extended diagonally upward. 
At 22 000 Ib., the crack extended almost to the bottom of the flange. 
The diagonal tension cracks were much larger than the tension crack 
in the middle portion. At 22000 lb., small horizontal cracks devel- 
oped at the level of the horizontal bar. At further loading, addi- 
tional horizontal cracks appeared. At failure, which took place at 
the loading of 28 600 lb., all previous horizontal cracks combined and 
formed a continuous crack extending from the support to the load, 
as is shown in the figure. 

For comparison, Fig. 13, p. 45, shows a typical tension failure of 
a beam of similar dimensions, as shown in Fig. 12, p. 40, but pro- 
vided with stirrups. All cracks are tension cracks and are confined 
to the center of the beam. ‘These two beams were selected from 
the series made by Bach. 


BEAMS WITHOUT SHEAR REINFORCEMENT 


The maximum unit shearing stress at which beams without web 
reinforcement fail by diagonal tension depends primarily upon the 
richness of the concrete and the age, and in smaller degree upon the 
percentage of steel and the ratio of depth to length of span. The 
last two items can be neglected in ordinary design. Since diagonal 
tension failure in beams without web reinforcement 1s sudden, a 
large factor of safety is necessary. (See p. 244.) In all cases quoted 


BEAMS WITHOUT SHEAR REINFORCEMENT 47 


below, unit shearing stresses were computed by Formula (63), 
p. 149. 

Effect of Age upon Web Resistance.—The effect of age is of 
great importance in determining the time for removal of forms and 
the age at which concrete can be loaded. It is illustrated in Fig. 14, 


p. 47, taken from tests made by Professor Talbot at the University 
of Illinois.! 


Unit Shearing Strength 
Pounds per Square Inch 





Age in Days 
Fig. 14.—Effect of Age upon Web Resistance. (See p. 47.) 


Tests by Pror. TALBOT 


Effect of Richness of Mixture of Concrete upon Web Resistance.— 
Figure 15, p. 47, taken from Professor Talbot’s ‘tests, shows the 
increase of web resistance with the amount of cement in concrete. 
The increase is quite marked although somewhat less than the 
increase in compressive strength. | 





1:3:6 Mix. 


Unit Shearing Strength 
Pounds per Square Inch 














Per Cent of Cement 


Fig. 15.—Effect of Proportion of Concrete upon Web Resistance. (See p. 47.) 
Tests by Pror. TALBoT 


Effect of Percentage of Horizontal Steel upon Web Resistance.— 
As is evident from Fig. 16, p. 48, the percentage of steel has a 


12 Bulletin No. 29, January 4, 1900. 


AS TESTS OF REINFORCED CONCRETE 


marked effect on web resistance, which can be attributed to two 
causes. First, for smaller percentages of steel, the deflection is larger 
with consequently increased tendency of concrete to crack. Second, 
with larger percentages of steel, the tensile stresses developed near 
the support are smaller, consequently the appearance of the tension 
cracks which later develop into diagonal tension cracks is retarded. 

















Vertical Shearing Strength 
Pounds per Square Inch 





90.75 1.00 1.25 a ee 2.25 


Per Cent Reinforcement 


Fic. 16.—Effect of Percentage of Horizontal Steel upon Web Resistance. 
(See p. 47.) 


Tests by Pror. TALBOT 


Ratio of Maximum Shearing Unit Stress Involving Diagonal 
Tension to the Modulus of Rupture of a Plain Beam and to the 
Compressive Strength.—In beams without web reinforcement, from 
tests by Professor Talbot,!® the ratio of maximum vertical shearing 
unit stress in beams failing by diagonal tension to modulus of rupture 
averages 0.5, and to the compressive strength, of 8 by 16-in. cyl- 
inders !* averages 0.09. 


BEAMS REINFORCED FOR TENSION AND COMPRESSION 


Tests prove conclusively the effectiveness of steel as compression 
reinforcement. 

Professor M. O. Withey’s Tests.!°"—The series of 1906 consisted of 
eight beams, 12 ft. long; breadth = 8 in.; height = 11 in.; depth 
to steel = 92 in., with 2.9 per cent tensile reinforcement and varying 
amounts of compressive reinforcement. The web reinforcement con- 
sisted of three bars bent up in two different places at a very flat angle. 


13 University of Illinois, Bulletin No. 29, January 4, 1909. 

14In determining this ratio the authors have converted the results found in 
cubes to a cylinder basis. 

15 Bulletins of the University of Wisconsin, Nos. 175 and 197, Series of 1906 
and 1907. 


BEAMS REINFORCED FOR TENSION AND COMPRESSION 49 


The results of the tests, although interesting, do not bring out 
fully the value of steel as compressive reinforcement, because all 
beams failed by diagonal tension, with the exception of the beam 
without compressive reinforcement, which failed in compression. 
Notwithstanding this, however, the maximum load of the beam 
without compression steel was 22 000 lb., while the maximum load 
for the beam with compressive reinforcement was 29 000 lb. 

The series of 1907 consisted of four beams, similar in design to 
the beams previously described, except that they were provided at 
each end with 10-j-in. round stirrups. All the beams failed in ten- 

sion at an average load of 34000 lb., showing an increase of 55 per 








Wis = Round 8 
7" / pees Round tp 
oot Fanon ten Es Sa ata Sh = 
H ae gu 
9a” er me 1, Round 
i VI - yr. Vill IX x, 


Fic. 17.—Dimensions of Beams, Stuttgart Tests. (See p. 49.) 
By Pror. C. Bacw 


cent over the beam without compression reinforcement. Still, because 
of the tension failure, the full value of compression reinforcement 
was not demonstrated. 

Bach’s Stuttgart Tests.!°—Bach’s tests of beams with compressive 
steel consisted of six types of beams, the dimensions and arrangement 
of reinforcement of which are shown in Fig. 17. The results of the 
tests are given on p. 50. The reinforcement of Beams VII, VIII, 
and IX is alike except that Beam VII in the middle portion has no 
stirrups while Beams VIII and IX have stirrups of the shapes shown 
in the drawing. The amount of the compressive reinforcement in 
Beam X is the same as in Beams VII and VIII, but the steel is of 
higher elastic limit. 


16 Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenieurwesen, 
Hefte 90 and 91. 


50 TESTS OF REINFORCED CONCRETE 


Test of Beams with Compression Reinforcement. (See p. 49.) 


Concrete: 1:3:4 by volume. Aggregates: Rhine sand up to 0.27 in. 
diameter, and Rhine gravel up to 0.79 in. diameter; 9.5 per cent of water by 
weight. Age of test, 45 days. 


Compiled from Tests by C. Bacu 


i 


Computed Unit Stresses in Steel and 
Concrete at Maximum Load 
Based on n = 15 








; és Compressive Baus 
Maximum Bicone Computed 
é Load Unit Unit Unit : Unit Fiber 
Specimens cre ie Cubes 
tensile |compressive; ’ stress Stress to 
lb. stress stress in Ib ._ | Strength 
in steel in steel concrete | — per'sd- 1"! of Cubes 
ts f's te 
lb. per lb. per lb. per 
sq. in. sq. In. sq. in. 
ie 16 860 19220 ») Mets eeu 2 220 1 590 1.40 
Aig ie 20 650 14 390 28 680 2 390 1 490 1.60 
Vite 27 500 17 750 29 200 2 500 1 480 1.69 
VIIL.. 29 000 18 720 29 800 2 670 1610 —-1.66 
iD. Ge 28 600 18 480 31 620 2 590 1.520 1.70 
2 e 36 000 23 200 | 36800 3 240 1 590 2.04 


Beams VI, VII, VIII, and 1X failed by compression. Beam X, 
in which the bond strength of the compressive steel was exceeded, 
failed at a considerably higher load than Beams IX, which had the 
same amount of reinforcement, because the compression steel in 
Beam X was of higher elastic limit. The table on p. 50 gives the 
maximum loads, the stresses in steel and concrete at the load under 
the assumption of n = 15, the strength of maximum cubes, and ratio 
of strength of cubes to figured stress in concrete in the beam. In 
Beams VII to IX, the compression steel reached its elastic limit first, 
and for further loading kept the same stress till the elastic limit 
of concrete was reached. In Beam X, on the other hand, the elastic 
limit in the concrete was reached first, and after this, stresses due 
to the additional loading were carried by the steel only until both 
materials reached the elastic limit. This points to the adjustment 
between compressive stresses in steel and concrete after one of the 


TESTS OF BOND BETWEEN CONCRETE AND STEEL ol 


materials passes its elastic limit. The same phenomenon was 
observed in the test of reinforced concrete columns. 

From inspection of the table, it is evident that for beams with 
compression steel, the theoretical unit stresses in extreme fiber in 
the concrete itself, computed for the maximum test load by formulas 
on p. 140 and on the basis of m = 15, are much larger than similar 
unit stresses at which the beams without compression reinforcement 
failed. Since the same concrete was used in all cases, it is rational to 
assume that this extra strength must be attributed to compressive ~ 
steel. This shows that the compressive steel carries larger stresses, 
and that its actual effect is greater, than would be expected from the 
formulas. It is especially noticeable in Beam X, for which the 
computed maximum fiber stress in concrete was 3 240 lb. per sq. in. 
while the crushing strength of the concrete was 1 590 Ib. per sq. in. 

The above tests prove conclusively that compressive steel may be 
relied upon to strengthen the compressive zone of a beam, and that 
its effect is even larger than would be expected from the formulas. 


TESTS OF BOND BETWEEN CONCRETE AND STEEL 


Bond between concrete and steel, or the resistance to withdrawal 
of steel imbedded in concrete, may be divided into two elements: 
(1) grip caused by shrinkage of concrete; (2) frictional resistance 
caused by the unevenness of the surface of the bar. Both elements 
act together until the bar begins to slip. Then the grip is destroyed 
and frictional resistance alone resists the pull. | 

In deformed bars, the grip and frictional resistance are aided by 
the bearing of the projections on the concrete, but this does not come 
into play until after the first slip. 

The pull-out tests are treated separately from the bond tests in 
beams, because the action of bond stresses is different in the two cases. 


PULL-OUT TESTS 


Pull-out test specimens consist of bars imbedded in blocks. The 
load is applied at the free end of the bar and is resisted by the resist- 
ance to withdrawal of the steel imbedded in the block. 

In practice, similar conditions occur in end anchors for fixed or 
cantilever beams where the concrete at the support corresponds 
to the block in the pull-out tests. ‘The maximum stress in steel at 


52 TESTS OF REINFORCED CONCRETE 


the edge of the support, which is transferred to the support by bond, 
corresponds to the applied force in pull-out tests. 

In computation, the bond stresses are considered as uniformly 
distributed over the whole surface of contact between steel and 
concrete. (See Formula 54, p. 268.) Actually, however, the bond 
stresses vary from a maximum at the edge of the support to a min- 
imum within the support. In many cases, in fact, the bar begins to 
slip at the place of application of the force before the bond resistance 
- of the whole bar comes into play. Therefore, ordinarily the portion 
of the bar near the point of support offers frictional resistance only, 
while the farther end of the bar offers erip and frictional resistance. 
The variation in magnitude of bond stresses along the length of | 
‘mbedded bar depends upon the length of imbedment. Hence, in 
basing allowable unit stresses on the tests, the effect of the ratio of 
the imbedded length to the diameter of bar must be taken into 
account. 

When a bar imbedded in concrete slips, the movement of the free 
end is somewhat greater than that of the imbedded end, the difference 
being equal to the deformation of the imbedded portion of the bar 
under stress. 

Effect on Bond Strength of the Ratio of Length of Imbedment 
to Diameter of Bar.—The average bond resistance, considered as 
_ distributed uniformly over the total surface area of imbedment, is 
smaller for long imbedments than for short imbedments. At the 
University of Ilinois,'” in tests by Mr. Duff A. Abrams for 1}-in. plain 
round bars imbedded in 1 : 2 : 4 concrete, seventy-four days old, the 
average bond resistance for 6-in. imbedment (4.8 diameter of the 
bars) was 420 Ib. per sq. in., while for 24-in. imbedment (19.2 diam- 
eters), it was 328 lb. per sq. in. Similar results were obtained by 
Prof. C. Bach.!® 7 

Method of Determining Bond Resistance.—In computing the 
bond resistance of a bar, the ratio of the length of imbedment to 
diameter of bar, and not the length of imbedment, is the determining 
item. The required length of imbedment increases in direct ratio 
with the increase of the diameter of bar. Thus a 25-in. imbedment is 
sufficient for a 3-in. bar because the ratio of the length to diameter is 
50. It would not be large enough for a 1-in. bar because the ratio 
then is only 25. (See p. 268.) 


17 University of Illinois Bulletin No. 71, December 8, 1913, p. 39. 
18(. Bach. Zeitschrift des Vereines Deutscher Ingenieure, 1911, S. 859. 


PULL-OUT TESTS ay) 


Bond Resistance for Different Slips.—Figure 18, p. 53, shows 
the relation between the bond stresses and slips for plain and deformed 
bars during the progress of loading. As is evident from this diagram, 
for plain bars initial slip occurred at 260 lb. per sq. in., or at about 
60 per cent of the maximum bond resistance. After the maximum 


700 








600 




















500 








1:2:4 Concrete 
Age about 2 Months 
= 











Bond Stress in Pounds per Square Inch 




















e 14 Deformed Round Bars, Embedment Variable 
+ 1 4 Plain Round Bars, Embedment Variable 
O Plain Round Bars, Diameter and Embedment Variable 





0 01 202 03 04 05 
Slip of Bar in Inches 


Fic. 18.—Relation of Bond Stress to Slip of Bar During Progress of Loading. !9 
(See p. 53.) 


bond resistance, which corresponds to a slip of 0.01 in. was reached, 
the resistance to withdrawal decreased. After a slip equal to five 
times the slip at maximum resistance has taken place, only 70 per 
cent of the maximum load is required to produce further slipping. 
The curves for the deformed bars are discussed on p. 54. 

Effect of Surface Condition and Shape of Bars.—The following 
conclusions may be drawn from Abrams’ tests: 


* University of Illinois Bulletin No. 71, December 8, 1913, p. 29. 


54 TESTS OF REINFORCED CONCRETE 


The bond resistance of square bars is only 75 per cent of the bond 
resistance of plain round bars. 

Rusted bars (with no scale) give bond resistance 15 per cent higher 
than similar bars with ordinary milled surface. 

The bond resistance of T-bars per unit of area decreases with the 
increase in size. For 1:2: 4 concrete, imbedment 8 in., and age 70 
days,2° the maximum bond resistance of 1-in. round plain bar was 
370 Ib. per sq. in.; of 1-in. T-bar, 310 lb. per sq. in.; and 2-in. T-bar, 
220 lb. per sq. in. 

Influence of Age and Mix.—The table on p. 58 gives the effect of 
age and mix on bond of {-in. plain round bars and of 3-in. corrugated 
square bars. py et 

Influence of Freezing—In Abrams’ tests, specimens made out- 
doors in freezing weather, where they probably froze and thawed 
several times during the period of setting and hardening, were almost 
devoid of bond strength. 

Ratio of Compressive Strength to Bond Resistance.—The ratio of 
bond strength at first slip to compressive strength, of 8 by 16-in. 
cylinder, is about 0.13, and of the maximum bond strength, 0.19. 

These ratios were determined by Mr. Abrams from tests on speci- 
mens varying in age from two days to 25 years, and proportions 
from 1:1:2+t01:5:10. These values agree very well with the 
results obtained by other experimenters. 

Deformed Bars.—Results of pull-out tests with deformed bars are 
given 2! on pp. 53, 55, and 57. ‘The first slip for the deformed bars 
occurs at about the same stress as for plain bars. 

After the first slip, the projections help to resist further slipping. 
Considering all the bond stresses except those resisted by frictional 
resistance taken by the projections, the bearing stresses on concrete 
for some types of deformed bars at large slips are very large, reaching 
in some cases 14 000 Ib. per sq. in. of the area in contact. This high 
compressive stress on concrete explains the splitting of the blocks in 
pull-out tests. Since the allowable working stresses are only a frac- 
tion of the ultimate bond stress the bearing stresses on projections 
always are within safe working limits. 

The maximum bond stresses, being accompanied by large slips, 
cannot be utilized in construction, where only a very small slip 1s 
permissible; consequently, the working bond stresses must be based 


20 University of Illinois Bulletin No. 71, December 8, 1913, p. 49. 
21 University of Illinois Bulletin No. 71, December 8, 1913. 


PULL-OUT TESTS 55 


Maximum Bond Stress and Bond Stress at 0.001 Inch Slip for Varying Propor- — 
tions and Ages.* (See p. 53.) 


By D. A. ABRAMS 





Stress in Pounds per Square Inch of Surface of Bar 





Proportions 


1) a AS ee Pera Sees: Tee an leo Lege .o 
Size of Bar Age 

Bue) 2 le] 8 jue] 2 ld) 8 le 

B(S2| 2 |Se| & |E5| 8 \Es| 2 Ss 

Rloe| & ISB) 2 [o8l 8 |o8) & \o8 

ie lew | elt le lak Viet | 


a ac | af a fn 


2 days PLOT HOT i2a! S123 rSol Salo) S2hr e277) a7 
4 days 197| 156] 231 | 195) 153} 110) 77| 43) 49) 32 
7 days 246} 202} 300 | 250} 226 | 158) 165) 112) 54) 32 
28 days 
to | 
32 days 


plain 60 ei 
round i | 530| 399) 554 | 492} 452 | 363) 311) 227) 190) 135 


65 ree 
120 days 
to 666] 479| 667 | 538} 603 | 469] 536] 398] 210} 172 
132 days 
16 months | 779] 656} 8961] 875} 8414] 800} 372| 333] 373] 253 
2 days 231| 96 205 | 92} 219] 97| 157| 44) 64| 13 
4 days 368| 1761 258 | 115] 305 | 129] 239] 74] 110] 23 
7 days 419| 171] 330 | 140} 459 | 187| 286| 104] 133] 35 
28 ae: 
: | 828| 344} 560 | 281} 641 | 306] 462| 179) 273] 97 
8-inch 
32 ee 
corru- 
ted 60 ibe 
ee | 1 132] 4981 1 053%] 556} 854 | 434] 623] 280] 391] 139 
square 


65 hee 


120 days 
to | 1 153] 599} 1070 | 564) 1079 | 576) 746) 326) 470; 159 
132 days 

16 months | 1 535) 892 


ailatb, shlewer| a eee ie Me (er ene re, 9) [pee ncwre: Is ee area a 8) 





| t 
ee 
* University of Illinois Bulletin No. 71, December 8, 1913, pp. 82-83. 
+ The reason for relatively low strength of Ll: 1: 2 concrete and 3-in. bars is unexplained 
and may be an erratic result. 
t Bars stressed to or beyond yield point. 


56 TESTS OF REINFORCED CONCRETE 


on stresses at a slip not exceeding 0.01 in. rather than on ultimate 
bond strength. The factor of safety for deformed bars based on this 
slip, however, may be made smaller than for plain bars, since the high 
ultimate bond strength and the existence of mechanical bond reduce 
the danger of actual bond failure. This is of special importance 
during construction, when comparatively green concrete may be 
called upon to support a considerable construction load. 

Allowable Working Stresses.—Allowable working unit stresses 
based on the tests are given on p. 263. 


BOND STRESSES IN BEAMS 


The method of computing bond stresses in reinforced concrete 
beams is given on p. 262. Although the formulas do not represent 
the actual conditions in a beam, they form, as explained below, a 
proper basis for design with values for working stresses based on tests 
and figured for the same assumptions. 

The computed maximum bond stresses in a beam occur at points of 
maximum shear. With uniform loading, this is at the supports and 
decreases uniformly to zero at the center of the beam. In beams 
loaded at one-third points, maximum bond stresses act in the outside 
thirds and are zero in the central portion of the beam. 

Phenomena of Bond Tests.—The bond stresses in beams are 
caused by the change from point to point, i.e., the increase, in the 
stresses in the longitudinal steel. This increase in stress in steel as 
computed is proportional to the amount of increase in the bending 
moment, and therefore equal to the vertical shear. Actually, how- 
ever, the change in stress in steel is affected by the presence of tensile | 
stresses in concrete, the amount and the proportion of which to the 
total tensile stresses is different in different parts of the beam. The 
effect is smaller near the point of maximum tensile stresses (where 
the concrete is cracked), and larger near the support where concrete 
may carry stresses even at maximum load. The increment of stresses 
in steel is not proportional to the shear; the bond stresses which are 
caused by that increment are, therefore, not proportional to the 
shear. 

The table on p. 57 gives observed bond stresses and computed 
bond stresses for varying intensities of loading for a beam loaded at 
one-third points. Since the shear between the support and the point 
of application of the load is constant, the computed bond stresses 


BOND STRESSES IN BEAMS 


Distribution of Bond Stress in Reinforced Concrete Beams. 


(See p. 56.) 


57 


Beams 8 by 12 in. in section and 10 in. deep to center of reinforcing bar. Loaded 
at the one-third points of a 10-ft. span. 


All beams failed by excessive tensile stress in the reinforcing bars. 


Compiled from Tests by Durr A. ABRAmMs * 


Beam | Size and Kind | Age at 


No. of Bar Test 


1055.6}One 1-in. Plain} 2 yr. 
Round 


1055.3)One I-in. Plain} 2 yr. 
Round 


1049 .3)One 1§ in. Cor-|13 mo. 
rugated 
Round 





Applied 


Beam 


Ib. 


2 000 


4 000 
6 000 
8 000 
10 000 
11 700 
2 000 
4 000 
6 000 
8 000 
10 000 
10 700 
2 000 
4 000 
6 000 
8 000 
10 000 
12 000 
14 000 
16 000 
18 000 
20 000 
21 000 
21 900 





Average 
Load on | Computed | Over region 
bond stress | just outside 


Ib. per sq. in. ies per sq. in.|Ib. per sq. in. 


38 

76 
114 
152 
190 
222 

38 

76 
114 
152 
190 
203 

34 

68 
102 
135 
170 
204 
236 
270 
306 
308 
359 


O10 








Observed Bond Stress 


of load 
points 


a) oye) Walt ie te” ae ie 


* University of Illinois Bulletin No, 71, December 8, 1913, p. 193. 
+ These stresses are, in general, the average bond stresses developed over a length of about 
12 in. in the portion of the beam about 4 to 16 in. outside the load points. 


t The average observed stress over a length of 9 to 151n. at the ends of the beam, 


16 
34 
36 
64 

Sly 

238 
15 
54 
95 

100 

130 

156 
20 
45 
95 

135 

150 

150 

225 

260 

290 

315 

360 

300 


Near ends 
of beam { 


58 TESTS OF REINFORCED CONCRETE 


are constant. ‘The observed bond stresses, however, near the sup- 
port are smaller than just outside of the points of application of the 
load until the steel reaches the elastic limit, after which a readjust- 
ment takes place and the bond stresses become equalized. From 
the table, it is evident, that in beams 1049.3, for example, the observed 
bond stress just outside of load points for a load of 16 000 Ib. is larger 
than the average computed bond stress for the ultimate load, 1e., 
21900 lb. This explains why, for beams failing by bond, the aver- 
age computed bond stress at the ultimate load based on Formula (50), 
p. 262, is smaller than the maximum bond strength in pull-out tests. 

The bond stresses given in subsequent discussion are those 
obtained by Formula (50), p. 262. 

Effect of the Distance of the Load from the Support on the Bond 
Resistance.—As may be inferred from the discussion of the phe- 
nomena of the bond stresses, the average bond resistance is larger as 
the load is placed nearer the support. Prof. C. Bach,” in tests of 
beams of 1 : 2:3 concrete at age of forty-five days, finds values of 
ultimate bond strength for distances 9.8 in., 19.7 in., and 29.5 in. 
from the support to average 507, 325 and 308 Ib. respectively. 








Bond in Beams 
in Pounds per Square Inch 














Age in Days 
Fig. 19.—Effect of Age on Bond in Beams 1: 2:4 Concrete. (See p. 58.) 
Tests by Pror. WITHEY 


Effect of Age on Bond. — Figure 19 on p. 58, from Professor 
Withey’s tests at the University of Wisconsin 7° shows the increase 
of bond strength with age, for 1 : 2 : 4 concrete. 

22 Widerstand Einbetonierten Eisens Gegen Gleiten. Einfluss der Haken, 


von C. Bach and O. Graf, p. 18. 
23 University of Wisconsin Bulletin No. 321, October, 1909, p. 27. 


BOND STRESSES IN BEAMS 59 


Professor Bach found for 1: 2:3 concrete the following bond 
strength: 


MR RO ee GY eee es 28 days 45 days 6 months One year 
Beams kept moist, lb. per sq. in... 278 308 393 435 
ca aa katy, is FO. afl 319 356 363 


He suggests the following formula for increase in bond strength 


with age: AM 
6 1 
(Fine 745(1 _ Vee :) 


Where uw = unit bond strength in lb. per sq. in; 
A = age in months. ) 


Effect of Mix of Concrete.—From tests, it is evident that the 
richness of mortar in concrete affects the bond strength considerably. 
The quality of stone is of little effect, provided pockets around the 
reinforcement are prevented. ‘The table below gives values for bond 
strength for concrete of different proportions. 


Bond Strength in Beams for Different Proportions of Concrete. (See p. 59.) 


Beams, 5 in. by 5 in. by 5 ft. 6 in. long. Reinforcement, 3-3-in. round bars. 
Lower bars imbedded in concrete for length of 10 inches at both supports. 
Beams tested on 5-foot span. Compressive tests on separate specimens. 


Compiled from Tests by Morton O. Wirury * 














: Age, | Average Bond | Compressive Strength 
es Days Bare jeerees te Ib. per sq. in. Ib. per sq. in. 
ap eae: 60 Limestone 276 1 790 
| 60 Gravel 215 2 200 
2326 60 Limestone » 216 830 
723 DOP eee SS 267 1 600 
| 





* University of Wisconsin Bulletin No. 321, October 1909, p. 27. 


Hooks as End Anchorage.—The requirements of a properly con- 
structed hook are: (1) it should permit the stressing of the steel to its 
elastic limit without appreciable movement; (2) the bearing stresses 
on the concrete must be within a safe limit. Since the allowable 
bearing stresses on concrete depend upon the properties of the con- 


60 TESTS OF REINFORCED CONCRETE 


crete, the factor of safety against crushing must be the same as that 
used in determining the allowable fiber stresses in concrete. Tests 
show that the crushing strength of concrete, when confined, is much 
larger than the crushing strength of cubes or cylinders. Hence, the 
safe bearing stress of the hook on the concrete should be based on 
the crushing strength of confined concrete. In comparing, there- 
fore, the relative efficiency of hooks, their bearing area is of first 
importance. 

When used for end anchorage, hooks which allow stressing the 
steel to elastic limit, but which at the same time split or crush the 
concrete, have not the required factor of safety as far as concrete is 
concerned, because at working stresses the concrete would have only 
a factor of safety of 2, instead of 3 as required by rational design. 

Tests 24 made for the Eastern Concrete Construction Company 
at the Massachusetts Institute of Technology determined the capa- 
city of the hook, but did not determine the load at which the first 
movement of the hook took place. . 

In all the tests, 2-in. round bars were imbedded in blocks 12 in. 
square and 15 in. long to a depth of 12 in.; with additional bends 
of different lengths. Right-angle bends and semicircular bends on 
a 3-in. diameter were tested. Several specimens of each type were 
tested, and the results were extremely uniform. 

The following conclusions may be drawn from the tests: 

(1) A 4-in. right-angle bend in a {-in. round bar (5 diameters) 
combined with 12-in. imbedment (16 diameters) is sufficient to stress 
the steel to its elastic limit. This hook, however, crushed the con- 
crete and split the block; therefore it does not give the required 
factor of safety against crushing of concrete. A longer bend does 
not increase the security because the bearing stress is not appreciably 
reduced. 

(2) A semicircular bend with a diameter four times the diam- 
eter of the bar is more effective than the square bend and is prefer- 
able because the bearings stresses on concrete can be kept within 
working limits. 

Action of Hooks in Beams.—Beams in which longitudinal steel 
is provided with hooks show a much larger load-carrying capacity 
than similar beams with ends of bars straight. Tests at age of 
forty-five days, by Professor Bach, on beams of 1: 2:3 concrete, 
12 in. square and 6-ft. span, reinforced with one 0.98-in. diameter 

: 24 Concrete, Plain and Reinforced, Second Edition, p. 466. 


SPLICES OF TENSILE REINFORCEMENT IN BEAMS 61 


round bar provided with three different kinds of hooks, gave the 
carrying capacity of the beam without hooks as 14 3380 lb.; with 
right-angle hook, 24 250 lb.; with 45° hook, 25 800 lb.; and with 

circular hook, 28 060 lb. The beam with Rak Sard neele failed 
by Pe htening the hook. 


SPLICES OF TENSILE REINFORCEMENT IN BEAMS AT POINTS OF 
MAXIMUM STRESS 


Tests have been made by H. Scheit and .O. Wawrziniok 2° to 
determine the effectiveness of different methods of splicing steel at 
the point of maximum stress. The beams were 12 in. square, of 
spans 63 ft. and 10 ft., reinforced with l-in. bar. They were tested 
with two symmetrical loads spaced 3 ft. 3 in. apart for the shorter 
beams, and 5 ft. apart for the longer beams. 

Straight splices were made with a lap of 10, 20, and 30 diameters 
respectively for the short beams, and 40, 50, 60, 70, and 80 diameters 
for the long beams; in hooked splices, the hooks consisted of a semi- 
circle with an inside diameter of 5 diameters of the bar and an extra 
length of 6 diameters of the bar parallel to the bar, and the bars 
were lapped 10 in, 20 in., and 30 in. respectively. 

Results.—For straight splices, the best results were obtained 
with a splice of 50 diameters, with which the elastic limit of steel 
was reached. 

Hooked splices proved very effective. Even a 10-diameter lap 
(the smallest lap used) in combination with a hook, as described 
above, was sufficient to provide the same carrying capacity as the 
beam without the splice, 


DEFLECTION © 


The deflection in reinforced concrete depends primarily upon the 
ratio of the depth of the beam, or slab, to the span. It also depends 
upon the percentage of tension and compression reinforcement, and in 
T-beams, upon the width of the flange. 

Influence of Percentage of Steel upon Deflection.—For equal 
depths and widths, the deflection of beams increases with the decrease 
in the percentage of tensile steel. Figure 20, p. 62, shows the 


25 Deutscher Ausschuss fiir Eisenbeton, Heft 14, 1912. 


62 TESTS OF REINFORCED CONCRETE 


deflections of beams 13 ft. long, 8 by 11 in. in cross section, tested 
on a 12-ft. span, by twe equal loads applied at one-third points. The 
test was made by Messrs. Richard L. Humphrey and L. H. Losse.*® 
The deformations in steel and concrete for the same beams are shown 
(oy 0s Ono 


20000 


78000 

















16000 Ce 
14000 - RRCRuGRaRecaa Oy ‘f 
12000 qt 


10000 : 1 


8000 
































ee 
































Total Applied Load in Pounds 





0.42 
6000 | 


















































4000 

















2000 






































0 0.05 010 0.15 020 0.25 0.30 0.85 0.40. 0.46 0.50- 0.55 


‘ Deflection of Beam in Inches 


Fic. 20.—Deflection of Beams with varying Percentage of Steel (See p. 61.) 
Tests by Ricuarp L. Humpurey and L. H. Losse 


Influence of Width of Flange upon Deflection.—In Bach’s tests *7 
to determine the effect of width of the flange, the results given in the 
table on p. 63 were obtained. It will be seen that although the 
percentage of steel based on the area of the stem is the same in all 
cases, the deflection for equal loads is smaller for beams with larger 
widths of flange. 


6 Technologic Paper No. 2, U. 8S. Bureau of Standards, June 27, 1911. 
27 Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenierwesen, 
Hefte 90 and 91. 


DEFLECTION 63 


Deflection of T-Beam with Varying Widths of Flanges. (See p. 62.) 


Span of beams, 9.84 feet; reinforcement, four 13%-in. round bars; load applied 
at one-third points. 
By C. Bacu * 





Deflection in Inches 





T-Beam, depth 9.84 in.; width of stem, 7.1 in. 























Total Load Rectangular | 
Beam Width of flange in inches 
7.1 X 9.84 in. | _ 

18.9 29.5 39.4 

lb. in. bar in. in. 
8 800 0.106 0.071 0.047 0.042 
17 600 0.376 Oe. 0.110 0.097 
ELE UO a a 0.368 0.188 0.161 
See MR ESS es a 0.290 0. 235 
2 AS ee Se ee 0.467 0.351 
rare Get Eck toto, Li Pas 0.544 








* Mitteilungen iiber Forschungsarbeiten aus dem Gebeite des Ingenierwesen, Hefte 
90 and 91. 


Influence of Compressive Steel upon Deflection.—The table on 
p. 64 gives deflection of beams without compressive reinforcement 
and with different percentages of compressive reinforcement. From 
the figures, it is evident that for equal percentage of tensile rein- 
forcement the deflection decreases with the increase of compression 
reinforcement. 


64 


Deflection of Beams with Compression Steel. 


TESTS OF REINFORCED CONCRETE 


(See p. 63.) 


Allbeams, 7.1 X 9.8 inches; span, 9.84 feet; tensile reinforcement, four 1,%;-in. 


round bars; load applied at one-third points. 
By C. Bacu * 


i 


Deflection in Inches 


eee nee ee ES Ee ee 





























Total Load Compression steel, per cent 
0 0.4 1.58 1 .58-% 
Ib. in. in. in. in. 
4 400 0.046 0.042 0.038 0: 037) 
8 800 0.119 0.102 0.086 0.084 
13 200 0.256 0.184 0.143 0.139 
17 BOO m4 Meohenra cree 0.298 0.210 0.203 
922-000. wl 52 Sa ee 0.287 0.276 
| 7 | 


* Mitteilungen tiber Forschungsarbeiten aus dem Gebiete des Ingenicurwesen, Hefte 90 
and 91. 
+ High elastic limit steel used. 


TESTS OF CONTINUOUS BEAMS. 


Since in concrete construction beams are usually continuous over 
several supports, it is of the greatest importance to determine by tests 
whether this continuity can be relied upon. 

Tests of Continuous Beams by Prof. H. Scheit and Dr. Ing. E. 
Probst.22—These tests included the concrete beams shown in Figs. 
21 to 23, pp. 66 to 68. Two beams of each type were tested to 
destruction. 

The spans and the reinforcement for the loaded spans were the 
same for all beams. The complete series included beams as follows: 

Type 1, simply supported beams. 

Type 2, continuous beams over two spans. | 

Type 3, continuous beams over three spans, end spans loaded. 

Type 3a, beams similar to Type 3, but supported by columns; 

loading same as Type 3. 
Type 4, beams of five spans, alternate spans loaded. 


28 “Untersuchungen an durchlaufenden Eisenbetonkonstruktionen,” Berlin, 
1912. 


TESTS OF CONTINUOUS BEAMS 65 


Figs. 21 to 23 show the cracks, and Fig. 22, Type 3, the deflection 
at different loads. Results of tests are discussed below: 

Comparison of Theoretical Deflection with Actual Deflection.— To 
determine the efficiency of continuous beams, the ratios of deflections 
of continuous beams to those of simple beams obtained from tests were 
compared with theoretical ratios for homogeneous beams. 

For continuous beams of two spans, Type 2, the observed ratio of 
deflections was between 0.42 and 0.45, while the theoretical ratio was 
0.40. For beams continuous over three spans with end spans only 
loaded, the observed ratio was 0.69 against the theoretical ratio 
of 0.74. 

For beams of Type 3 of the same design as the one above but in 
which the ends were connected with columns, the ratio varied from 
0.34 to 0.37. From the theoretical figures, it appears that as far as 
deflection is concerned, this type is almost midway between a beam 
fixed at one end, for which the ratio is 0.40, and a beam fixed at both 
ends, for which the ratio is 0.20. 

Type 2.—The beams continuous over two spans failed at the sup- 
port at an average load of 14 240 lb. per lin. ft. The theoretical 


2 
negative bending moment at the support is — = The stress in 


steel at the support for the maximum load, figured on the basis of 
the above bending moment, is about 64000 lb. per sq. in. It is 
evident that this stress is much higher than the elastic limit of the 
steel used in the test, which shows that the assumed theoretical 
bending moment coefficient is too large. Based on the moment of 
resistance for the yield point of steel, we get a bending moment 
coefficient of 10 instead of the theoretical 8. The point of inflection 
was found to coincide with the theoretical point of inflection. (See 
Fig. 21, p. 66.) 

Type 3.—In the beams continuous over three spans, the first cracks 
appeared at the bottom of the beam in the central portion of the loaded 
span at a load of 2 565 lb. per lin. ft. In the unloaded span, the first 
crack appeared at the top of the beam at a load of 5 600 lb. per lin. ft. 
The sequence of other cracks and the loads at which they appeared is 
evident from the illustration. The failure at a load of 12 600 Ib. per 
lin. ft. was caused by passing of the elastic limit of steel in the loaded 
spans. 

The deflection diagram, Fig. 22, p. 67, gives a positive proof of 
the continuous action of the beam. As is evident from the diagram, 


66 TESTS OF REINFORCED CONCRETE 


the deflection in the end span is positive, while in the center span it is 
negative. The computed stresses and bending moments at the dif- 
ferent loads agree quite closely with the measured stresses. 

The measured compressive stress for the maximum load in the 
middle span (which was not loaded) was found to be 1 500 |b. per sq. 
in., which is almost identical with the theoretical stress. The cracks 
in the unloaded span, which are uniformly distributed over its whole 
length, furnish a conclusive proof of continuous action. If no pro- | 







t u a 
ie. Hits 72 
. J ‘ 4 \, 


Position of Loading 


5l'Round 
°76 oun 





SSUES 
JINN. La 
Teo! 71" 5-10" Round km 3'- 10" phe 9 
ee ivebeecseneeee Qin.) e=neeane man ane na 






SS SN as 
51/9 3h 37X 3/ 
| i haat 
4 G2) 9G) 4) BYIG) NC5) 
A 
Cracks in Beam During Loading 


Fic. 21.—Continuous Beam of Two Spans; Type 2. (See p. 65.) 


vision had been made for the negative bending moment in the 
unloaded span, failure would have been certain. 

Type 3a.—In the beam continuous over three spans and mono- 
lithic with columns, as shown by Fig. 23, p. 68, the connection 
between beams and columns was not rigid, as would be the case with 
rigid frames, but was built as in ordinary building construction. The 
beams were of exactly the same design as in Type 3. The compari- 
son, therefore, gives the effect of the connection of the beam with the 
column. The first cracks in the loaded span appeared at a load of 
4590 Ib per lin. ft., and in the unloaded span, at a load of 10 935 lb. 
per lin. ft. The corresponding figures in Type 3 were 3 565 lb. per 
lin. ft. and 5 670 lb. per lin. ft. At a load of 13905 Ib. per lin. ft., 


TESTS OF CONTINUOUS BEAMS 67 


the first cracks appeared at the top of the end column; and at 16 740 
Ib. per lin. ft., a crack appeared at the top of the middle column. 
The beam failed at 17 620 lb. per lin. ft. by steel passing the elastic 
limit. At the time of failure, cracks were observed in the com- 
pressive part of the interior column. 


foes Ie go 7d pith Ao Idec, » 78 
OS ae a ae fg tn a 





3-5" Round 10-5" Round 



















pe 7" r ane 11" y" 
She aie ae, ie 2 8G one (396 Round ot Round -----~ < 
ES) ‘, ry @ \TN 
a A. 2-9 et y we aN : x ati : : a0 5 i x 5 
: ree nf 17g | Rou i git = pe ee - cf 
tz iz Section ae-7f =; ?, ees : pak 8 = 10- 6 & a : 
a-b ee oe re eaten sp 7 md 0 cogsee carton ieee ae Tae ee H 
g.! "pound '5- Up ae a, 
16 ound Arrangement of Steel 
2 -H Round” 





0.098" Deflection of Beam 9.098" 


Fic. 22.—Continuous Beam of Three Spans; Type 3. (See p. 65.) 


After the tests, cracks were found at the bottom of the columns, 
located in reverse position to the cracks at the top. During test, 
not only the beams, but also the columns deflected, which shows that 
the whole construction acted as a unit. The deflection in the beams 
was smaller than in Type 8, as explained before. 

As was expected, the moment of resistance at the ultimate load 
does not agree with the bending moments for continuous beams based 
on the assumption of free ends. The construction must be con- 
sidered as a frame. Dr. Probst finds that the positive bending 


68 TESTS OF REINFORCED CONCRETE 


moment coefficient in the loaded span was 12.02, which agrees very 
closely with the bending moment coefficient computed by him by 
the rigid-frame method. 

Type 4.—The cracks in the T-beams continuous over five spans 
indicate clearly that the beams acted as continuous. From the 









" 
6 -2 Round 3 - "Round:, 
xe 4 3" 7 saat ”" “ 
J jl g Round 2-H Round... See ol Round-.,, 3-4 Round--.. 
“ 45.3" Round & Pate ee aft ae 
\ 8 y} oe “ o\ X ba 2 16 Round “Sek ‘ 
i ES. Y 





——_—_—_——" 


COTA 















9" 






















‘ } a fey 
¢ I$ -F aS --S--t -2->| 11" jx -+-4 
; c 48 y 74-76 Round | 
ay c \ \ Xf Val, 47 Nal Vy V) V 7 YIN a? 
vi HT NNN ANAT PUN 
~--l.5 AN = d f ‘ 
1 Ley, SV SOE 4 - Ea cil be i 1 17 ' 
pape Ci i S ke aT : nephe p yi/ gym (5-76 Round 
pa AY yop Roundy] |S Ma A ROU gs 
4 2 ai an ‘ ae ee ead ‘ 
J 8, Rounds 7 oats pouea / 4 6 Round’ "Pe Tp ia, 6 - H'Round 
“5 Round“ "5 ound ‘7-7 Round 
Section on c- d Arrangement of Steel 
Pe hep ned pt pete Rw ps tere tk BO. So Me ee phere oe lu 
Bol} hes : Teag ¢ "I <5 144 >| aa 
















" 


[one oc aenedee 2] Oot=noenon-- Shee anne nes G1 Gi--so eee bf Qiasansn ies 
Position of Loading 


49 49 

49 +9 7 62 62 62 
49 4949 

62 23 31 26 





20 26 23 
23 20 


Cracks During Loading 


Fic. 23.—Continuous Beams of Three Spans; Type 3a. (See p. 66.) 


comparison of the moment of resistance of the beam at the maximum 
load, with the theoretical bending moment obtained from ordinary 
continuous beam formulas, we find a very close agreement, which 


proves that even for five spans a continuous beam acts as contin- 


uOUS. 


DISTRIBUTION OF CONCENTRATED LOAD ON SLABS 69 


DISTRIBUTION OF CONCENTRATED LOAD ON WIDE SLABS 


A concentrated load placed on a wide slab is carried not by the 
portion of the slab immediately below the load but by a considerable 
width of the slab. Experiments described below, conducted on slabs 
where the maximum width was equal to double the span length, show 
that within these limits the whole width of the slabs is affected by a 
concentrated load placed centrally, and, therefore, assists in larger 
or smaller degree in carrying the load. Thus, in a 32-f{t. slab tested 
on a 16-ft. span, the whole slab was affected by the load. This con- 
dition is brought about by the shearing resistance of concrete and not 
by any distributing reinforcement. 

Distribution of Stresses and Deflections.—The stresses and 
deflections in the center af the span are largest in the section of slab 
directly under the load. For sections away from the load, the center 
stresses and deflections decrease rapidly at first and then more slowly, 
until for wide slabs they become insignificant near the edges of the 
slab. 

Definition of Effective Width of Slab.—As explained above, in a 
wide slab under a concentrated load, along the width of the slab, 
there is considerable variation in stresses in the center of the span. 
The ordinary slab formulas assume uniform distribution of stresses 
along the whole width; therefore they cannot be used directly without 
some assumption. In order to be able to use ordinary slab formulas 
for the design of wide slabs loaded by concentrated loads, it is neces- 
sary to determine the width of slab of same depth which, stressed uni- 
formly along its whole width, would have the same strength as the 
wide slab with its varying distribution of stresses, and for which the 
moment of resistance, based on an accepted maximum unit stress, 
would be equal to the moment of resistance of the whole width of 
slab with a varying distribution of stress and with largest stress equal 
to the maximum accepted stress. This width is called effective 
width. It is less than the total width of the slab; and the ratio of 
effective width to total width depends upon the ratio of the total 
width to the span. 

Effective Width from Tests for Central Load.—From the tests 
made by the Office of Public Roads and reported by A. T. Goldbeck,?® 
the ratio of the effective width of slab to the span depends upon the 


* A. T. Goldbeck. The Influence of Total Width on the Effective Width of 
Reinforced Concrete Slabs Subjected to Central Concentrated Loading. Pro- 
ceedings, Am. Concrete Institute, 1917. Vol. XIII, pp. 78 to 88. 


70 TESTS OF REINFORCED CONCRETE 


ratio of the total width of the slab to the span. The ratio may be 
found from the table below. 


Effective Width of Slab for Concentrated Load 


eee 


\| 
| 





Total Effective see Total Effective erste a 

Width Width “ Width Width a 

of Slab | of Slab Wate oe “f Slab |. OR eal eae 

Total Width |, Total Width 

Gain 0.11 1 Lait 0.671 0.61 
0.21 0.21 1 Lot 0.68 1 0.57 
0.37 0.281 0.93 {34 0.701 0.54 
0.41 0.371 0.93 1°44 0.711 0.51 
0.51 0.441 0.88 | 1.51 O78 0.48 
0.61 0.501 0.83 1.61 0.721 0.45 
0.71 0.551 0.79 L7H 0.721 0.42 
0.81 0.581 0°72 1.81 0.721 0.40 
0.91 0.621 0.69 1.91 0.721 0.38 
ot 0.65 1 0.65 2.01 0.721 0.36 





1 = span of the slab. 


From the above table it is evident that, where the width of the 
slab is equal to 0.4 of the span, practically the whole width is effective. 
When the width of span equals 1.5 1, the effective width attains its 
maximum value, namely 0.72 1. In such cases, about one-half of 
the total width is effective. The increase of the total width beyond 
14 times the span has no effect on the effective width, as in all cases 
the effective width equals 0.72 U. 

From the tests made by Professor C. T. Morris, the effective 
width equals 0.6 times span plus 1.7 ft. for a width equal to 1.351 + 4 
ft. This checks closely enough the value in the table. 

Effective Width from Tests with Eccentric Load.—When the load 
is not applied in the center of the width of the slab, the following rule 
may be used.®° 

If the distance from the load to the nearer edge of the slab is 
greater than one-half the effective width of the slab, considered as 


30 See: Tests of Large Sized Reinforced Concrete Slab Subjected to Eccentric 
Concentrated Loads, by A. T. Gotppeck and H. 8. FairBANK. Journal of 
Agricultural Research. Vol. XI, No. 10, December 3, 1917. 


DISTRIBUTION OF CONCENTRATED LOAD ON SLABS ik 


centrally loaded, the effective width may be assumed to be same as for 
centrally loaded slab. 

If the distance from the load to the nearer edge of the slab is less 
than one-half the effective width of centrally loaded slab, the effective 
width may be taken as one-half the effective width for centrally 
loaded slab plus the distance of the load to the nearer edge. 


Example 1: Find effective width for a slab 20 ft. wide with a 14-ft. span, 
when the concentrated load is placed 3 ft. from the edge. 

Solution: The ratio of total width to span is 2¢ = 1.43. Fora slab loaded 
with a concentrated load placed centrally, the effective width, from table on 
p. 70, is 0.711 = 0.71 XK 14 = 10 ft. The distance of the concentrated load 
from the nearest edge is less than one-half of this width; therefore the effective 
width for eccentric load is 

49 +3 = 8 ft. 


Description of Tests.—Tests were made by the Office of Public 
Roads and Rural Engineering on the following slabs: 




















Reinforcement 
i Span | Width Ne eae aan eo meer eomerre BESS nig ae 

Size bars Seti sh 

it ft in. in. in. 
835 16 oa 12 103 3 in. sq. 10 001d 
930 16 oo 10 8i hte ae 8.87 O275 
934 16 a2 fe 6 eee GanG O75 
A19 6 12 7 6 mek | inne ORT 
A20 6 12 4 3 es 6.26 i ii 
AQ1 6 12 5 4 Ae esta 0.75 
A22 6 oe 16 15 Seen. | 5 OP 75 
A23 6 12 6 5 eA | 6.66 0.75 

| 


The mix of concrete in all slabs was 1: 2:4. No cross reinforce- 
ment was used. 

All slabs were tested with their original width. Slab A25, after 
being tested with its original width, was gradually reduced in 
width by having successive sections split off one side. 

Method of Test.—Concentrated loads were applied in the center 
of the slab. Additional tests were made with two and four con- 
centrated loads. In some slabs the loading was carried to destruc- 
tion; while in slabs tested with different widths up to the working 


72 TESTS OF REINFORCED CONCRETE 


loads, deformation of concrete and deflection readings were taken 
at all widths. The deformation readings on concrete were made 
perpendicularly to the support at regular intervals along the whole 
width of the slab. In several cases, deformations were also taken 
longitudinally with the span. Steel deformations were also meas- 

ured in some slabs. 
Conerete deformations were found to be a most satisfactory 
measure of distribution of stresses in the slab. In interpreting the 
results, however, it was 

































































































































































necessary to take into 
omge mi Sat account the flow of con- 
Es Effective Width 49 erete under — sustained 
= ero loads. . 
3+ Co Slab Width =10.0 L Steel deformations | 
= es Effective Width+ 628-1 were less satl sfactory, 
= ro T cH because the readings at 
SAE Y TCE Slab Width=-8'0 He - points where cracks oc- 
=e ai Effective Widtht6.24-| curred were different from 
ic 7 Sa) iG readings at adjoining 
ret | H is points without cracks. 
[ ce Slab Width =4'0 Deformation of Slab 
1 Effective Widih}372_| and Effective Width.— 
iastaiek Figure 24 shows the © 
Slab Width- 2:0 deformations of steel in 
ae el Effective Width}1.s3_| Slab A19 when tested 
Feo : | with different widths. 
Span 6' Effective Depth 6. Steel 0.75 per cent The effective width is 
Fic, 24.—Steel Deformations and Effective marked by dash lines. 
Width Slab A19. (See p. 72.) Results of Tests.— Fig- 


: ure 25, p. 73, shows the 
results of all tests. In the figure, the ratios of total width to span 
are plotted as abscissas and the ratios of effective width to span as 
ordinates. The curve represents the values given in the table on 
p. 70 and suggested for use. To be on the safe side, the curve is 
made to coincide with the lowest values obtained in the tests. 

Breaking Loads of Slabs.—Slabs 835, 930, and 934 were loaded to 
destruction. The results are given on p. 73. | 


DISTRIBUTION OF CONCENTRATED LOAD ON SLABS 73 





Breaking Loads of Slabs. (See p. 72.) 














; Reinforcement Breaking 
Effective eu Hh emt, Lopate 
Number penn Load 
EP, p | Size | Spacing lb. 
835 10} 0.0075 | fin.sq. | 10.5in. | 119.000 
930 83 0.0075 | aa Ne | poy evi 80 000 
934 6 fer ODOT irs ats S tery. Oe We peas OO? 40 000 


All slabs were 32 ft. wide and 16 ft. span. 


Infliénce of Total Width on Effective Width of Reinforced Concrete Slabs Subjected to Concentrated Loading. 



































































































V7 ; 
Beet ttt 
7.0 
bet nal ae 
0.8 
es 
Se nol] ae 
Sls 05 ++ Fy 0A-19, é'Span, 6" Depthe 
SI es x A-20, 3 H 
a4 ee ee ee Note: Symbols like those in eet AER eet, ie ae 
0.3 ol left column are based on FAR-R2 16 Te via? 
ae | ys as ain those 
& S27 ae in right on steel deformations. 
0.1 
ma «| 


s iw © 

So 0G) 94S) Sl Re SS es Se 
Total Width 
Span Length 


Fig. 25.—Influence of Total Width on Effective Width of Reinforced Concrete 
Slabs Subjected to Concentrated Loads. (See p. 72.) 


a 

' 

' 

' 

' 

1 

' 

’ 

' 

’ 

° 
& 
for) 
=) 
ah 


Gage Aeaey 





74 TESTS OF REINFORCED CONCRETE 


It will be noticed that the breaking loads for different effective depths 
of the slabs are in the same relation as the squares of the depths of the 
slabs. This indicates that the effective width for all depths was 
practically the same. 

Fig. 26, p. 73, shows the manner of failure of a slab under a 
concentrated load. 


TESTS TO DETERMINE DISTRIBUTION OF LOAD BY SLAB TO JOISTS 


The tests show that if a continuous concrete slab is supported 
by several parallel joists of any material, and a concentrated load is 
placed directly above one joist, the load is distributed by the rigidity 
of the slab to several joists (see Fig. 27). The distribution depends 
upon the ratio of the thickness of the slab to the span. 


100 
90 


80 


Per Cent of Load Carried by Middle Beam 





Ratio of Slab Span to Thickness 


Fig. 27.—Distribution of Slab Load to Three Parallel Supporting Joists. 
(See p. 74.) 


TESTS BY PROF. C. T. MORRIS 


The laboratory test made by Professor C. T. Morris consisted 
of slabs, 6, 7, and 8 in. thick respectively, supported on three lines of 
10-in., 25-lb. I-beams spaced 3 ft. 6 in. on centers. The span of 
the joists was 12 ft., and they were supported on other I-beams, 
which in turn rested on concrete pedestals, as in bridge construction. 
The load was placed directly over the middle joist at its center. 

In distributing the load, the slab acts as a cantilever supported 


DISTRIBUTION OF LOAD BY SLABS TO JOISTS 75 


at the point of application of the load and loaded by the reaction on 
the side. 
The following conclusions may be drawn from the tests: 

(1) The percentage of reinforcement in the slab has little or no 
effect upon the load distribution to the joists, so long as safe loads 
on the slab are not exceeded. 

(2) If the span is ten times the thickness of the slab, or more, total 
load must be considered as carried by the joist under the load. The 
amount of load distributed by the slab to other joists than the one 
immediately under the load increases with the thickness of the slab. 

(3) The outside joists should be designed for the same total live 
load as the intermediate joists. 

(4) The axle load of a truck may be considered as distributed uni- 
formly over a 12-ft. width of roadway. 

Fig. 28, p. 75, shows the elongation in extreme fiber of the steel 
beam and deflection for the middle beam and for outside beams. 


0.120 
0.100 


0.080 


on in Inches 


' 0.060 


Deflecti 


0.040 


Deformation in Divisions on Strain Gage 





Z 


veal 
“ae 
HA | 
ee 
Zara 


0 
0 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 


Load in Pounds 
Fre. 28.—Elongation in Extreme Fiber of the Steel Beam and Deflection of 
the Middle Beam and of Outside Beams. (See p. 75.) 


Value of One Division on the Strain Gage is 0.00019 in. 


Similar results would be obtained with concrete joists. 

The conclusions apply only to cases in which ratio of span to 
thickness of slab does not exceed 10. The largest ratio used was 7, 
but the results may be exterpolated. 


76 TESTS OF REINFORCED CONCRETE 


TESTS OF PLAIN CONCRETE COLUMNS 


Professor Talbot 3! made a very comprehensive series of tests, the 
results of which are given in the table below. Conclusions, con- 
firmed also by experimenters abroad, are as follows: 

(1) Manner of Failure.—Plain columns fail either by shearing at a 
diagonal plane of fracture, or by crushing, when the material is shat- 
tered and cracked longitudinally. The diagonal shearing failure 
almost always occurs suddenly and with little or no warning, while 
the compressive failure 1s more gradual. 


Tests of Plain Concrete Columns. (See p. 76.) 


Materials: Portland cement, Wabash River sand, crushed limestone. Col- 
umns, 12.in. diameter, round, 10 ft. long. 


By Pror. Arraur N. TaLBot 


See 


nee 














rue ae 
Average |Maximum Y aneeeh) Minimum * 
ae cee we 
Number Broparkorns Average Slee Ultimate Pee Conk Ultimate Be Ga 
of of Age Unit Unit. techy Unit j 
Specimens Strength] Strength Strength eh 
Concrete Average Average 
Tested 
noo Ib. per | lb. per o% lb. per of 
Y* |) sqcimteslasedoan. : sq. in. a 
|__| mee sae 
2 1 Lear 64 2 300 2480 Ges 2 120 — 8 
7 125) 2844 65 1 740 2 210 at 1165 —33 
2 shearers 4 9) 613 1 033 1110 ts 955 — 7 
2 14 28 63 575 575: alse aa ee 575 
6 122.314 192 2 025 2 680 32 1 770 —13 
2 Pls 2B 801 1d moe 2 0 2770 


: 
trees | 


(2) Effect of Richness of Concrete.—The strength of columns 
increases in nearly the same proportion, 1.e., almost as a straight line, 
with the increase in the proportion of cement to total dry material 
used. (See Volume II.) 

(3) Modulus of Elasticity and Poisson’s Ratio.—The modulus of 
elasticity of the columns is almost constant for the first third of the 
strength of the concrete. Beyond this point the modulus decreases 
till it reaches at the ultimate load about one-half of its initial value. 


31 University of Illinois Bulletin No. 20, 1908, 


TESTS OF PLAIN CONCRETE COLUMNS fire 


The Poisson’s ratio, or the ratio of the lateral to the longitudinal 
deformation (see Vol. II) wasfound for 1 : 2 : 4 concrete to be between 
0.10 and 0.17 up to a load of about one-half the ultimate. It 
increases with the load, reaching probably 0.25 at the ultimate 
load. 

(4) Effect of Repetition. Repetition has no effect on deformation 
for loads up to one-half of the breakirtg strength of the column. For 
higher loads, the deformation increases after repeated applications of 
the load. After ten repetitions of a load equal to three-fourths the 
normal breaking strength, for example, the deformation was increased 
by 25 per cent. 

It must be noted, moreover, that the suddenness of failure of plain 
concrete is increased by the length of the column. This absolutely 
excludes plain concrete columns from structures where they are apt to 
be exposed to shock or to secondary stresses due to bending, as in 
building construction. | 

Concrete vs. Brick Columns.—Tests carried out by the U. 8. 
Bureau of Standards on columns of common, hard, and vitrified 
brick laid with lime and cement mortar, indicate that the strength 
varies with quality of brick and mortar, while large and small columns 
- show about the same unit strength. 

A series showing the strength of piers of common, hard, and vit- 
rified brick, laid with different mortars, is given in the following table. 
The lime mortar specimens showed an almost entire lack of carbona- 
tion on the interior. Three piers of each kind of brick and mortar 
were made, with headers every other course, every fourth course, 
and every seventh course, but this variable appeared to have no 
effect. Bricks were laid flat. 

Two large-size columns 48 in. square and 12 ft. high, of common, 
hard-burned brick, one laid in 1 : 1 cement mortar and one in 1: 3 
lime mortar, were tested by the Bureau and crushed at 2920 and 
760 Ib. per sq. in. respectively.?? 

Tests made at the Watertown Arsenal and quoted by the Com- 
mittee of the American Society of Civil Engineers, on the Compressive 
Strength of Cement,?? give the ultimate strength of common brick 
piers about eighteen months old as ranging from 800 to 2 400 Ib. per 
sq. in., the results for brick laid with lime mortar averaging nearer 


32 James E. Howard, Engineering Record, March 22, 1913, p. 332. 
8§ "Transactions, American Society of Civil Engineers, Vol. XV, p. 717, and 
Vol. XVIII, p. 264. 


78 TESTS OF REINFORCED CONCRETE 


the lower figure, and those for 1 : 2 Portland cement mortar nearer 
the higher figure. 


Compressive Strength of Brick Piers * 


Tests by the U. 8. Bureau of Standards. (See p. 77.) 


Dimensions 30 inches square by 10 ft. high 


a een aN I eee ee 

















Nes Compressive 
Ixind of Brick Mortar Bes Strength, 
Months : 
Ib. per sq. in. 
ah Os (| 1:3 lime Oiben':: 170 
Pe gee tad cae \ | 1:3 cement 1 575 
(| 1:6 lime 4 910 
15% lime 
Hardy uta een oe ee rae (aap ene 1 1465 
1 : 3 cement 1 1650 
(| 1:6 lime 4 1360 
anh (15% lime 
Viviicd to ee ig Ee : ae 1 2.900 
1:3 cement 1 2780 


| 
| | 


* Engineering News, August 5, 1915, p. 242. 


The unit stresses allowed by the New York Borough of Man- 
hattan Building Code, 1916, for brickwork are, 


Brickwork in: Ib. per sq. in. 
Portland cement: Mortar... 4.<...<,.) ose ee 250 
Natural cement mortar.......... 1 ies 210 
Lime cement mortar .2.0. -... «. 2-5 eee 160 
Lime mortar... 22.0 Jo. es oe ee 110 


The first value is but little more than one-half that recommended 
for good 1 : 2: 4 Portland cement concrete. 


TESTS OF COLUMNS REINFORCED WITH VERTICAL STEEL 


Tests prove positively that in reinforced concrete columns, steel 
and concrete are effective in resisting the load carried by the columns. 
As explained in the chapter on the Theory of Reinforced Concrete, 
p. 159, the stress in steel equals the stress in concrete multiplied by 
the ratio of their respective moduli of elasticity. This fact also 1s 
borne out by the tests. 


TESTS OF COLUMNS VERTICALLY REINFORCED 79 


Mr. Spitzer,** in Austria, and Professor Withey,*® at the Univer- 
sity of Wisconsin, observed that, near the ultimate load, adjustment 
between stresses in steel and concrete takes place, so that finally the 
failure occurs by both of the materials passing the elastic limit simul- 
taneously. ‘This adjustment may be explained as follows: 

Suppose that one of the two materials reaches its elastic limit, 
but there still exists bond between them. If the load on the column 
is increased, there is a tendency for the increase to be distributed 
over both materials. However, any increase in stress, in material 
which had passed the elastic limit, would produce very large defor- 
mations, which the other material, being still within the elastic limit, 
could not undergo. Since both of them must deform by the same 
amount, this other material takes all the stresses due to any increase 
of the load, till it finally reaches its elastic limit, and the column fails. 
The table on p. 81 gives results of tests of full-sized columns made at 
the Watertown Arsenal. 

Manner of Failure.—Contrary to expectations, in most cases 
failure in columns occurs near the top or bottom instead of at the 
center. ‘This has been explained as probably due to greater porosity 
of concrete at ends. 

In most columns, hair cracks appeared at 85 to 90 per cent of the 
maximum load. In some cases, however, the failures were sudden. 
In a number of cases, concrete split at the column reinforcement 
after its elastic limit was reached. The splitting effect is caused by 
the lateral deformation of steel, which exerts sufficient pressure on 
the concrete to break it. The steel, therefore, should be placed at a 
sufficient distance, say at least 1 in. from the face of the concrete. 
With proper protection, there is no danger of buckling till after the 
elastic limit of the steel is reached. 

From Spitzer’s tests, it would appear that columns with steel 
placed well within the cross section of the column are somewhat 
stronger than those with steel placed near the surface. In practice, 
however, columns are apt to be subjected to eccentric loading; there- 
fore, the placing of steel in abnormal positions must be discouraged. 

Factor of Safety.—A column with vertical steel only is liable to fail 
without notice when its ultimate strength is reached. The ultimate 


4 Mitteilungen Uber-Versuche ausgefiihrt vom Eisenbeton-Ausschuss des 
osterreichischen Ingenieur und Architekten-Vereins, ‘“‘Versuche mit Eisenbeton- 
saulen,” Heft 3. 

> University of Wisconsin Bulletin No. 466, December, 1911. 


80 TESTS OF REINFORCED CONCRETE 


strength of concrete columns, even if built under the same conditions, 
is more variable than that of steel columns. Therefore, the commonly 
accepted factor of safety is larger than that used in steel columns. 
When designed according to formulas given on p. 406 with allowable 
unit stresses on p. 407, columns with vertical steel only are very 
reliable. 

Modulus of Elasticity for Reinforced Columns.—The modulus of 
elasticity varies for different intensities of loading and for different 
mixes of concrete. In selecting the ratio of moduli of elasticity to be 
used in design, it is proper to be euided more by the required factor 
of safety than by the actual modulus of elasticity at any particular 
stage of the loading. From the tests thus far made, the moduli of 
elasticity given on p. 407, with the suggested working stresses, give 
the required factor of safety. 

Rich Versus Lean Mix.—As is evident from the table on p. 81, 
cement is very good reinforcement for the column, as the increase in 
strength is much larger than the cost of additional cement. 

Influence of Bands.—In tests made by Mr. Spitzer,2° of columns 
having different spacing of bands, those in which the spacing was 
equal to, or smaller than, the diameter of the column gave somewhat 
greater strength than columns in which the spacing exceeded the 
diameter of column. 

Influence of Percentage of Steel.—The tests show clearly that 
the effect of reinforcement in columns is the same whether the per- 
centage is large or small. In all cases the steel takes a stress equal to 
the stress in concrete times the ratio of moduli of elasticity. 

Watertown Arsenal Tests.—The table on page 81, from test by 
Mr. James E. Howard at the Watertown Arsenal, gives the relation of 
actual tests to theoretical computations based on a ratio of elasticity 
of 15. It is noticeable that the actual strength is almost always more 
than the theoretical, and this is especially the case with the leaner 
mixtures because the modulus of elasticity of the leaner concrete 1s 
lower, and therefore the ratio of 15 is very conservative. 

An excellent analytical treatment of columns reinforced with ver- 
tical steel is given by Professor Talbot in one of his University Bulle- 
tins37 The problem is discussed briefly by one of the authors in a 
paper before the Boston Society of Civil Engineers.?% 

36 See footnote on page 79. 

37 University of Illinois Bulletin No. 12, F eb. 1, 1907. 


38 Sanford E. Thompson, in Journal Association Engineering Societies, June, 
1907, p. 316. 


TESTS OF COLUMNS VERTICALLY REINFORCED Si 


Many of the tests at the Watertown Arsenal, for example, were 
made with vertical bars imbedded in columns 12 in. square and 8 ft. 
Jong, with absolutely no bands or horizontal steel of any kind placed 
around these vertical bars to hold them in place; that is, the bars 8 
ft. in length were placed in the four corners of the column—in some 
tests only 2 in. from the surface—and simply held in place by the 2 
in. of concrete." There was no sign whatever of buckling until 
the compression was so great that the elastic limit of the steel was 
passed. 


Strength of Plain Concrete and Mortar Columns vs. Vertically Reinforced 
Concrete 


Columns 12” & 12’. Height 8 feet. Age of Mortar and Concrete 6 months, 
Watertown Arsenal. (See p. 80.) 





Reinforced Columns 
Proportions| Plain 








Concrete 
or Reinforcement Computed | Reference 
Mortar strength | to ‘Tests 
Columns Actual | based on |of Metals” 
Actual Ratio |strength|column (4)| U.S. A. 
= Strength area | lb. per | and aratio 
a et ib:iper Description | steel to} sq. in. | of n = 15 
a) e 3 Sq. in area lb. per 
o soiy+y 
O}2|R column Sq. in 
(1) |(2) |(3) (4) (5) (6) (7) (8) (9) 
1;2]0j} 3070 | 8-2” round bars | 0.029 | 4200 4290 |1905, p. 377 
3. 1:0 2380 | 8-2’’ round bars | 0.029 | 3840 3320 = |1905, p. 377 
1);4/0 1520 | 8-2’ round bars | 0.029 | 3380 2120) (1905, p. 377 > 
Ph 1 0 1080 | 8-2’ round bars | 0.029 | 2810 1510 = |1905, p. 377 
24-51) 0 1080 |13-2” round bars | 0.046 | 3 900 1780 = =|1905, p. 377 © 
1 | 1 | 2*| 1720 | 4-2” twisted bars} 0.014 | 2.890 2060 |1904, p. 386 
1 | 2] 3*| 1769 | 4-3” twisted bars} 0.014 | 2010] 2100 1904, p. 386 
1/2/)4)] 1418 | 4-0.74"X*0.74” 
trussed bars} 0.014 | 1900 1689 —_|1906, p. 538 
2 | 4*| 1710 | 4-2’ twisted bars| 0.014 | 1990 2050 = |1904, p. 386 
2|47T| 2400 | 8-2” twisted bars} 0.029 | 3700 3360 = |1907, p. 242 
1/|3{|6 | 1450 | 8-3” corr. bars 0.019 | 2290 1840 = ‘|1904, p. 379 


1906, p. 535 


* 1-in. to 14-in. pebbles. 
7 Age 17 months 22 days. 


39 Test of Metals, U. S. A., 1905, p. 344. 


82 TESTS OF REINFORCED CONCRETE 


TESTS OF SPIRAL COLUMNS 


In analyzing results from tests of spiral columns, it is necessary to 
examine not only the strength, but also the shortening of the column. 
In building construction, because of the dependence of the different 
members upon each other, it is advisable to permit a shortening, or 






























































Stress in Pounds per Square Inch 


a 

















1 Per-Cent 
s Spiral Steel 
Ges | Ly | I (ae 
S - AI ; - | 6.1 Per-Cent 
tis Vertical Steel 
j 
0 
S S S S S S S S S S [~) S S S S S 
MD one ou ee ae Say Se eye es mates 
a ees SS im cae ee 


Deformation per Unit of Length 


Fic. 29.—Deformation Curves for Spiral Columns with Varying Amount of 
Vertical Steel.4° (See p. 82.) 


Concrete 1 : 2 : 34; 1% Spiral High Carbon Steel; 0% Vertical Mild Steel. 
Concrete 1 : 2 : 33; 1% Spiral High Carbon Steel; 3.8% Vertical Mild Steel. 
Concrete 1 : 2: 34; 1% Spiral High Carbon Steel; 6.1% Vertical Mild Steel. 


deformation, of not more than 0.007 of the length of the column. 
The factor of safety, therefore, must be based on the load causing a 
certain deformation rather than on ultimate strength. Although the 
strength of column beyond the yield point is not available in ordinary 
construction, the greater ultimate strength, ductility, and uniformity 
in strength of spiral columns reduces the danger of sudden failure, 


40 University of Wisconsin Bulletin No. 466, December, 1911, pp. 92, 98, 94. 
By Morton O. WiTHEY.* | 


TESTS OF SPIRAL COLUMNS 83 


and larger unit stresses may be allowed. In practice, it is more 
rational to increase wcrking unit stress in concrete than to compute 
the stress by a formula which takes into account the steel in the 
spiral. (See p. 421.) | 

Harly European tests of spiral columns were made on short 
columns and without deformation diagrams; and as a result, very 
high working stresses based on the ultimate strength were recom- 
mended. More recent tests, notably by Talbot and Withey in the 
United States, showed excessive deformations of spiral columns at 
high stresses. 

Column Tests by Prof. M. O. Withey.—The tests at the Univer- 
sity of Wisconsin, Series of 1910, consisted of four series, two of which 
will be considered below; namely, series 1, columns with varying 
percentage of longitudinal and lateral reinforcement, and series 2, 
columns with varying proportions of concrete. The table on p. 84 
gives the general results of the tests. 

The following general conclusions may be drawn from the tests: 

1. The cheapest way of increasing the strength of a column is by 
using a rich mix. 

2. Spiral reinforcement greatly increases toughness and ultimate 
strength of a column, but does not raise the yield point. (See col- 
umns M and O, p. 84.) The strength beyond the yield point cannot 
be utilized in building construction; hence, the amount of steel 
for spirals should be made only large enough to produce required 
ductility and raise the factor of safety against failure. In practice, 
1 per cent of spiral reinforcement seems to be sufficient. | 

3. Longitudinal steel increases the stiffness of the column and 
raises the yield point. 

4. Stress in steel at the yield point of columns is practically the 
same for all mixes of concrete and only a little below the yield point 
of the vertical steel. (See table, p. 85.) This phenomenon may be 
explained by the fact that for leaner concrete the ultimate strength is 
smaller, but the deformation at the yield point larger than for rich 
mixes. For rich concrete, the stress in concrete at the yield point is 
larger, but the ratio of the moduli of steel to concrete decreases. 
Since the stress depends upon the product of stress in concrete times 
the ratio of the moduli of elasticity, the two values evidently adjust 
themselves so that the product is the same in all cases. 

5. Columns loaded eccentrically give results which agree closely 
with the formula given on p. 175, as is evident from Fig. 30, p. 86, 


84 TESTS OF REINFORCED CONCRETE 


Tests to Determine the Effects of Varying the Percentage of Vertical 
By Morton 


Average values given. Columns C-1, 2, 3, 4, D-1, 2, 3, 4, 120 inches long; all others 


| | 



































2 Reinforcement Par Cent ss 
5 Reinforce- to 
4 Spiral com & 
ie) iw 
ates Area Maxi- 2 
Cc ] i) Ta a mum & 
olumn s a ore ES 
Number nb z | a8 A Vertical Load - 
<q je bars, 
Ss . round Wire | Pitch Veree Spiral P 
es cal 
lal A 
EP | | P |lb. per 
d jc4 in | lb. sq. in 
—— ee LD ————— 
Series I. 
fe NA 2a, eee eee a ——— 
| | | | | 
wit-3|1:214 | 52| 8.2} 86.6 | 0 | A pies eit bd 0 | 225 500 | 2600 
4-1, 25) 1 ee 85) BT Sra Te ee No.7| 2” | 0 | 0.50 | 175 500 | 22385 
C1, Bl Apt 2487 533] 8.5 TR. O18 a Nowe aie 2.00 | 0.50 | 259 500 | 3 300 
11, 23 bile Ss 57 | 8.5| 78.5 | 8 a’ | No. 7 GAY 3.78 | 0.50 | 327 000 | 4 160 
Pe ae as: Ses BS] 85) 7820 1-8 VO Now & ne 6.11 | 0.50 | 402 750 | 5 120 
Telos el boes 34| 573 potas Say SU No. 7 1G 0 1.00 | 207 450 | 2 640 
1-1, 2 el aoaeh oa 57.1 8.5). 78.0 | 3 TEN Oa ie 2.00 | 1.00 | 306 750 | 3905 
Neto 1 23 Ss 574| 8.5 78.518)" 4" No. 7 1 3.78 | 1.00 | 329 250 | 4 195 
M-1,.2:)) 222.533 574| 8.5 Veena || ce) a” | No.7 Ve | 6.11 | 1.00 | 368 250 | 4 685 
Pal,.2) 128 583| 8.5| 78.9 Sci eee aN Ookg ake 8.00-| 1.00 | 543 900 | 6 925 
Cyto Pe see te B71 RIS) T8tb 4S uu a” ht 6.11 | 12:96 | 516-750 6 580 
Pele 2 |Piecees 89)1 S578 4 Sel a a” a 8.00 | 1.96 | 547 500 | 6 965 
Q-1, 24 bt Zee 4 | 53 | 8:5) 78.5 | 8 13” ar" ge j10. 12 | 1.96 | 556 500 | 7 090 
CLS 12 4 1 6Se LOOP 7S aS F eee nee a" | A 0 | 2.00 | 316 750; 4 030 
(Ladue lec en G04} 10 0] 78.5 | 9 g/” | a" i 3.900 | 2.00 | 373 250 | 4 750 
1 ! \ | 














Series II. 
Series th. 
| | | 
TA, BA La 483| 8.3| 82.6 | 8 a” Re ag 5.83 | 1.00 | 443 500 | 5 370 
X1/ 2 11 2 1E 333)" aGe Sei 8o i001 0 No. 7 ie 0 1 OO Ne ee ties eae oe 
V2 2 (Li 1epess Giles 18s26. is 4” No. 7 1s 5 75 id OOM wacyae ceame ren 
Soto ty Lol see eee Us israel 6 No. 7 see 0 1.00 | 460 500 | 5 855 
Pepe 2a Le ae 561| 8.5| 78.5 | 8 av VT Nott ES 6.11 | 1:00} 513 000 | 7 290 
Vote belesdecd 563| 8.5| 78.5 | 0 No. 7 1” 0 1.00 | 420 000 | 5 340 
U-i, 21.1: 128 561| 8.5| 78.5 | 8 Bh NOick ne 6.11 | 1.00 | 640 500 | 8 150 
Asie ls icaa | ee, 8.3) 84.6 (°8 9 3") No v 1% 1.86 | 0.99 | 617 500 |7 3007 
AB-1, 2| 121.33. | 593 8.3| 83.6 | 8 24” | No.7 1 3.55 | 1.00 | 617 500 |7 3907 
AC-1,' 2) 131.33.) 53% g 9|.84:6 18.4% | Nog 1 5.69 | 0.90 | 617 500 |7 4807 
AD-1, 2] 1:21.33 1195 | 8:2) 85.0 Soe TN Omer ii 5 62) 41 OL OSTiaR se wess cae Or | 
| 





WAY eee ee 
* University of Wisconsin Bulletin No. 466, Engineering Series, Vol. 7, Now 
























































TESTS OF SPIRAL COLUMNS 85 
and Spiral Reinforcement in Reinforced Concrete Columns. (See p. 83.) 
O. WiTHEY* 
102 inches. 
a Stresses at Yield Point * 2g Elastic Properties 
a, | Bb 
A. g |e 
os In steel 2 & ee me Es 
Load fi a ve oO I Modulus; 2 SB 
at as In In ‘ie poo 8 of & | nis 
Yield | 2 | core con- | @0 he Elasticity) ° | .» | 84 
Point | £3 | crete | 6° | 2A at 4 2 Bete et 
ene PE ees Beeline Ulti- 5 a | #2 
3s | ertical} Spira 5 is SO oe = aoe eR 
So Py S28 Strength| «= e 5s 
eo —— % BAA 8 mr P 3 
a A fart} fos hee ad abil Von a 
, & 2 | > 
Py Pi | Ib. per | Ib. per | Ib. per jIb. per|lb. per fe Ib. per eben ae 
lb. P sq. in. : Sie iD Chin. | bq. in. i sd.in. ) jf. sq. in. | myiRy | my{ny | 
| | | 
ant ae oe a oe eala aout ele OOO OOOM 182 6 ree et, 
im 5000.88.)" 1855 Wsoe..%. 10 050 | 1855 | 1750 | 0.94 |2 350 ee 1320 127.0" 10, 127 
213 000} 0.83 | 2710 | 40500} 12000; 7940 |} 1760] 0.91 |3 350000; 11.0 | 21.0 '0.110 
273.000) 0.84 | 3470 | 42 000 6 450 | 1960} 2180] 0.13 |3 400000) 13.0 | 22.0 |0.110 
333 000} 0.83 | 4240 | 40650; 6150 1 860 | 2055} 1.11 |4 350000} 11.5 | 22.0 |0.088 
HOS000) OF 92) 1 80 Poi... 4 800; 1370} 1770) 1.29 |2 100.000] 14.5 | 19.5 |0.146 
205 500} 0.67 | 2615 | 39450} 7050) 1 865|.1995} 1.07 !2 800030, 13.5 | 21.5 |0.123 
265 500} 0.81 | 3375 | 38700} 6900] 1985} 1800} 0.90 |3 800 000] 11.0 | 20.0 |0.130 
295 500; 0.80 | 3760 | 37 650 6 450 | 1565 | 1 685 | 1.09 |3 425000] 18.5 | 24.0 |0.101 
445 500! 0.82 | 5660 | 39150; 8100); 2750] 2 365] 0.86 |5 300000} 9.6 | 14.5 |0.121 
348 000; 0.68 | 4430 | 36900) 8400} 2310] 2 480/ 1.08 |4650000) 9.9 | 16.0 eae 
408 000) 0.74 | 5190 | 37 500 8 250 | 2 385 | 2390] 1.005)5 250 000, On We 16). OF Oal2s 
453 000} 0.82 | 5765 | 37-200 6 750 | 2 230 | 2 305] 1.03 |5 500000] 10.8 | 17.5 |0.094 
Ber ene. | 2385. ). sae. lee ee Doon | cate OOM a0 s OS eels aes mene oe ete heya en coene te flea comets 
280 ia 0.75 | 3 575 | AQSBOO WR. ate « 2 2238 | 2 252 | LO eas Meeasy tcl has 2 trea: DRM SiN ie care 
| | saet 
264 500| 0.60 | 3200 | 36900) 8 250] 1115} 1770 | 1.59 |3950000|12.80 |...... 0.085 
338 000)...... alrooue ds. 2). 8 400 | 4100 | 4760} 1.16 |5050000| 5.95 |...... 0.175 
470 000)......}] 5700 | 30600}; 6300] 4160] 4120] 0.99 {6 700000} 5.70 |...... 0.190 
318 000} 0.69 | 4050 |....... 7 350 | 4050 | 4070] 1.01 |4125 000} 7.30 |...... 0,124 
453 000} 0.79 | 5760 | 37050! 7 350| 3725 | 4400] 1.18 {5 175000) 8.45 |...... 0.140 
PSU OOO ONG T to G70 fon. ck. 6.750 | 3.570 | 4 880 | 1.87 |3 400000; 8.80 |...... 0.135 
468 000! 0.73 | 5950 | 36450] 6750] 3970/ 4555] 1.15 |5 250000] 8.40 |...... 0.125 
489 000) 5 Sao 5780 | 34750 | 7050) 5185) 6 787 | 1.80 |5625 000; 5.90 |...... 0.190 
BLE OOO} ss 0.. 6 150 | 35250} 7700] 5090} 6 387] 1.25 |5 950000} 5.95 |...... 0.185 
BA2'000)'-. .. .. 6 590 | 36000} 6600) 4760] 6075] 1.28 |5 150 000) 6.40 |...... 0.180 
RE Pe Te eel ete y aisle js. sis, |a)4 o2e0% «| a a waite s [> 543000 (HOO ne eerie Oe oes 


+ Column did not fail. 


86 TESTS OF REINFORCED CONCRETE 


in which the straight lines represent figured stresses in steel and con- 
crete by Formulas on pp. 175 and 176, while the dots and circles 
show the actual stresses obtained from deformation. 

Action of Columns under Test.—Up to the point of breaking 
strength of plain concrete, the action of the columns with spirals 
was the same as for columns with vertical steel only. ‘The observed 
stress in spirals was from 6 000 to 8 000 Ib. per sq. in. For spiral 
columns with vertical steel, the deformation curve continues as a prac- 
tically straight line to the yield point of the column. The yielding 
is indicated by the scaling off of the protective shell and by an 






2000 


‘Average Unit Stress 
in Pounds per Sq. In 






| ae 


| 


Bay Eccentricity 1 4 in. 


He 0 2000 4000 6000 8000 10000 0 2000 4000 6000 8000 
ua 0 10000 20000 30000 40000 50000 0 10000 20000 30000 40000 
: Unit Stress in Concrete or Steel in Pounds per Sq. In. 


5.96 Per Cent of Vertical Steel 0.97 Per Cent of Spiral Steel 






nN 


, i 
if Uh ie 
nN 


Fic. 30.—Comparison of Theoretical and Actual Stresses for Eccentric Loading.* 
(See p. 85.) } 


‘ncrease in ratio of lateral and longitudinal deformation to the 
applied load. ‘The yield point is more marked for columns with 
large percentage of reinforcement (see Fig. 29, Pp. 82). 3 Lor 
spiral columns without vertical steel, the deformation diagram is a 
curve without a marked yield point, so that the yield point is only 
distinguishable by scaling of the shell. 

After the yield point has been passed, the disintegration of the 


41 University of Wisconsin Bulletin No. 466, December, 1911, p. 71. 


TESTS OF SPIRAL COLUMNS 87 


shell progresses very rapidly. The ratio of shortening due to the 
applied load becomes larger and final failure takes place by buckling 
of the column, or, in columns with a small amount of lateral steel, 
by breaking of the spirals. 

Stresses in Steel and Concrete.—The table on p. 84 gives the 
stresses in steel and concrete at yield point and at maximum load. 
The stresses in vertical steel were obtained from the deformation by 
using a modulus of elasticity of 30000000. The remainder of the 
load assumed as carried by the concrete, and divided by the area of 
the core, gave the unit stress in the concrete. In figuring the stress 
in concrete, the area of the core was used in preference to the total 
area of the column, because at yield point and at maximum load, 
either a part or the whole of the outside shell is destroyed and is then 
ineffective for carrying the load. 

The stress in spirals obtained from lateral deformation is very 
small at the yield point of the column, which corroborates the state- 
ment that up to yield point 4? the spirals do not affect the column 
appreciably. 

The table is of interest in giving a comparison of the stresses in 
steel with the stresses in concrete for different mixes and also in giving 
the values of the ratio of moduli, n. Although the value of this ratio 
was variable, the stress in steel at the yield point of the column was 
about the same for all columns, which seems to show that in a column 
under load an adjustment of stresses takes place. From the table, 
also, it is evident that the stress in concrete at the yield point was the 
same irrespective of the amount of vertical reinforcement. 

An interesting experiment was tried in connection with tests by 
Wayss and Freytag? The columns, after reaching the maximum 
stress, were again loaded after nine months and after one year, and 
showed a large increase of strength over the original maximum 
strength, in some cases reaching 50 per cent increase. After these two 
loadings, the column spirals were removed and the core tested again. 
In no case was the core, upon removal of the spirals, found to be dis- 
integrated; in fact, each core showed a considerable strength, which 
tends to disprove a contention previously held by several authorities 
that the concrete in hooped columns becomes disintegrated after a 
certain point in loading is reached and is simply prevented from 
flowing by the hoops. 


42 See also Professor Talbot’s Bulletin No. 20, University of Illinois. 
43 Morsch, 4th edition, p. 117. 


88 TESTS OF REINFORCED CONCRETE 


TESTS OF SQUARE COLUMN WITH RECTANGULAR BANDS 


Tests were made by Wayss and Freytag 43 on eleven types of 
square and rectangular in cross section with closely 
spaced hoops and bands. These tests prove that the bands in square 
columns are not very effective and can not be compared as to effi- 
ciency with spirals in round columns. The outside shell started break- 
ing off as soon as the concrete reached its maximum crushing strength. 


TESTS OF COLUMNS WITH STRUCTURAL STEEL REINFORCEMENT 


The size of column can be reduced by the use of structural shapes, 
rigid enough to serve as 
a structural steel column, 


12-in. columns, 


a = om imbedded in concrete. 
Below are given results 


from tests of columns with 
two types of structural 
steel. The results must 
not be considered as apply- 
ing to all steel sections and 
must be used with caution 
where the structural mem- 
bers differ materially from 
those in the tests. 





_--- Outline of 
spiraled type 


_16 inches” 


Ge 





gar ler Outline of 

H core type 

ON. yor Outline of 
dh bes 5 fireproofed type 





11, or 14 spaces. 


’ NOTE 


a” ” 
Angles: 2% 24x+ 
fie sLRIKO Et 
ea coe} z 


5 au . 
Rivets: 3 diameter 





Fig. 31.—Dimensions of Test Columns. ** 
(See p. 88.) 


43 Morsh, 4th Edition, p. 117. 


Talbot-Lord Tests of 
Columns.’—The Talbot- 
Lord tests consisted of 
thirty-two columns dl- 
vided into four groups: 

1. Plain steel columns. 

2. Core-type columns, 
i.e., columns in which the 
portion within the struc- 
tural steel members was 
filled with concrete. 

3. Fireproofed columns, 
i.e., core-type columns 
having a 2-in. protective 
covering. 


44 University of Illinois Bulletin No. 56, Fig. 1,-ps 4.4 
45 University of Illinois Bulletin No. 56. 


TESTS OF COLUMNS WITH STRUCTURAL STEEL 89 


4, Spiraled columns, i.e.; core-type columns enclosed in close- 
fitting spiral and filled with concrete to outer surface of spiral. 

The cross section of the structural steel, which is shown in Fig. 31, 
p. 88, was the same for all columns. Ratio of length to minimum 
radius of gyration varied from 6.1 to 59.5. The results of the tests 
are given on p. 90. 

Plain Steel Columns.—No bending was visible to the eye at the 
maximum load. After the maximum load was passed, bending 
developed very gradually. The average deformations per unit of 
length are shown in Fig. 32, below. ; 





400000 








200000 


Applied Load in Pounds 


100000 





0 ,0002 ~=.0004 .0006 0008 .0010 0012-0014 0076 .0018 
Deformation per Unit of Length 


Fig. 32.—Average Deformations of Plain Steel Columns.‘* (See p. 89.) 


The effect of the length of the column was more marked at high 
than at low loads. According to Professor Talbot, the ultimate 


stress in column for different ratios of ! may be represented by a 


Beet tine.tormuls, = = 36 500 — 155 : 

Core-type Columns.—The columns of core type were very tough 
and failure was slow. For short columns, the failure was caused in 
most cases by crushing of the concrete; for longer columns, by bend- 
ing and crushing of concrete. No bending visible to the eye was 
observed until maximum load was reached. The effect of mixture 


46 University of Illinois Bulletin No. 56, 1912, p. 19, 


90 


TESTS OF REINFORCED CONCRETE 


Tests of Steel Columns, Reinforced with Concrete 


By Tausor and Lorp.* 


Age of concrete, 59 to 61 days. 


(See Fig. 31, p. 88.) 





























P.S. = Plain steel column of 8 — 3’ X 23’ X 3” angles with no concrete core. 
GC. T. = Core-type column of 8 — 3” angles with concrete core. 
F. = Fireproofed column same as core type with 2” of extra concrete outside of 
structural steel. No spiral. 
S. = Spiraled column same as core type with spiral steel and concrete core 
filled out to conform with diameter of spiral. 
= 5 Stresses in 
3 « |‘82°8| Total Load in lb. 
sn al ee lb. per sq. in. 
=e 5 |eeor 
E rah ope 
ia Se NS B 2s. 
je) 
Column | 0 15 aaa ae ir an Load | Load 
ae Con- ogs 
Number os ci ae con- con- 
‘, crete | & |-="6 3 : In 
5/8 2 ls0e Column! sidered | sidered | In a 
aha) Hm load | carried | carried | steel 
1 b i; ,crete 
sq: : x y 
im. r steel {concrete 
Lite a ee eee ee —_—_|—_—_—— EN ee eee omen 
ft. in : 

8 902 P.S.| 13} none | 2-0) 6.1 | 487 300). Oke eee BL DOON ey wc. 
8 905-6 P.S.| 13} none | 4-8} 14.4 |444600).......]....... 34200), 221. 
8 907-8 G.T 12011 22 24) 4-8in 589 500) 444 600} 144 900) 34 200] 1 355 
8 910-11-14|P.S. | 13} none |10-0| 30.8 | 420 000).......|....... 32 400]. ..2. 
8 912-13 “1 CrT 201 28s 410-0) eee 547 350| 418 000] 129 350) 32 150} 1 210 
8915-16 |P.S.| 13} none |15-4| 47.2 |372 000).....-7).. 22. 28 GUO = ser 
8 917-18 © |C.T.120)1 224/154) 500 350| 372 000} 128 350] 28 600} 1 200 
8 920-21 |P.S.| 13} none |19-4| 59.5 | 359 600|....... Sige hee 27-B501 As 
§ 922-238 (CATA120/1 224110 4) ee 493 450] 359 600] 133 850} 27 650) 1 250 
8'925-26. 1G. T. 120i tlees2 110-0 645 500] 418 000} 227 500} 32 150] 2 125 
8 027-28 | (Oc 12011 S55 O11 O30 ae vies 523 250! 418 000) 105 250) 32 150) 985 
8 930-31.) JF Se 4213102 2410) eee 633 200] 418 000} 215 200} 32 150) 1 075 

8 933 S.7 \153i1 22s A0-O18 aa 600 000} Applied 5 times; not broken. 

8 934 Sit 115311 32 24/100) 856 000) Not broken; near ultimate. 

8 935 Sit [P53ite 22 LOO a 600 000] Applied 3 times; not broken. 
Sire good wechcang | eA eae ae ree leet ce 830 000] Second test; near ultimate. 

8 936 S. 4 (15302 4 10-0 ene. 625 000; Not broken. 

8 937 S/t ise Ie 200 uae): 830 000 

| oh: Pam el eae 714 000} Second test with spiral and 





* University of Illinois Bulletin 
+ 0.75% of spiral reinforcement. 
t 1.00% of spiral reinforcement. 





outside concrete removed. 


No. 56, 1912, pp. i4 and 105. 


TESTS OF COLUMNS WITH STRUCTURAL STEEL | il 


of concrete on the strength was small, because the strength of the 
column was governed by the steel rather than by the concrete. 

In Fig. 33, p. 92, are shown deformations per unit of length 
of core-type column. Deformations of plain steel columns are also 
shown for comparison. (See also Fig. 32.) | 

The relative loads carried by the concrete and by the steel, 
respectively, are given in the table on p. 90. The amounts were 
determined by assuming that for equal deformations the structural 
shapes in the core-type column carried the same load as a similar 
plain steel column and the balance was carried by the concrete. 

Fireproofed Columns.—The behavior of fireproofed columns of the 
core type was about the same as that of the column without fireproof- 
ing. The concrete shell outside of the structural shapes, however, 
remained intact until the ultimate deformation of the column was 
nearly reached, but its effective unit strength was lower than the 
unit strength of the concrete core, probably because the shell failed 
before the maximum stress in steel and concrete core -was reached. 

Even if the protective covering is not relied upon as adding to the 
strength of the column, it is advisable to tie it by means of hoops or 
spirals of large pitch so as to prevent spalling in case of fire. 

Spiraled Columns with Structural Steel—The core-type column 
with surface spirally reinforced exhibited larger strength and tough- 
ness than similar columns without spiral. The thin protective cover 
remained intact. The strength of the columns exceeded the capacity 
of the Illinois and Lehigh University testing machines, which is 
830 000 Ib. 

Because of the large deformation, the increase in strength afforded 
by the spiral is not available in ordinary building construction; hence 
a large percentage of spiral is not justifiable. One per cent of spiral 
reinforcement, however, makes the column tougher and safer and 
also prevents the outer shell from spalling. Since the danger of 
sudden failure is removed, such columns may be designed with some- 
what higher working loads than are allowed for the fireproofed type. 

Tests by Professor Withey.‘7—The structural steel reinforcement 
consisted of four angles 2 < 2 X 3; in. placed in the four corners of a 
square column. The out-to-out dimensions of the steel core were 
8 in. square. The result of this test agrees with the tests previously 
described. 


47 University of Wisconsin Bulletin No. 300, ‘‘Tests of Plain and Reinforced 
Concrete Columns,” 


92 


600000 


600000 


400000 















300000 


200000 


coe 


100000 


4 Numbers 8907 - 8 


500000 


Applied Load in Pounds 


400000 
300000 
200000 


100000 


/ 





Numbers 8917 - 18 


0 0005 0010 .0015 0 








TESTS OF REINFORCED CONCRETE 











Length 10'-0” 





Numbers 8912 - 13 


Length 19'- 4" 


Numbers 8922 - 23 


0005. ~.0010.~—=«w00785 


Deformation per Unit of Length 


Fic. 33.—Load Deformation Diagrams for Plai 
ing Core-type Columns.** 


Norr.—Curves labeled Column refer to core ty 
steel type. See also Fig. 89. 


48 University of Illinois Bulletin No. 


n Steel Columns and Correspond- 
(See p. 91.) 


pe; those labeled Steel refer to plain 


56, March, 1912, p. 24. 


TESTS OF LONG COLUMNS 93 


Recommendations for Design of Core-Type Columns.—<As a 
result of this test, the ultimate strength of the core-type column may 
be considered as consisting of the strength of the plain steel column 
plus the strength of the concrete core figured with a unit stress equal 
to the strength of concrete in cylinders. 

Recommendations for the Design of Fireproofed Columns.— 
Tests by Professor Talbot and Professor Withey show that in columns 
with structural steel the protective cover does not fail until the col- 
umn reaches its maximum load. In practice, the protective cover 
is not considered as adding to the strength of the column. There- 
fore, in designing this type of column, the recommendations given in 
connection with core-type columns may be used. Similar results 
were obtained by Emperger and Spitzer.*® 


TESTS OF LONG COLUMNS 


Tests by Spitzer on columns, 9.8 by 9.8 in. cross section, 9.9 ft., 14.8 
ft., and 23 ft. long respectively, with ratio of length to the last diam- 
eter of 12, 18, and 28, show no appreciable difference in strength 
and no buckling in any of the columns. 

To determine the effect of slenderness, Professor Bach ®° com- 
pared the strength of 4 ft. column (ratio of slenderness 4.3) with that 
of a 29.5 ft. column (ratio of slenderness 32), and found the strength 
of the longer column to be 0.75 of the strength of the short column. 

As a result of this test, Professor Bach suggests the following 
formula for the strength of a long column in terms of the strength of a 
short column, which in turn may be assumed equivalent to the 
strength of 8 X 16-in. cylinders. 


1 
‘Eew eh AP. (1) 


aa Terie 
1 + 0.0072 7 





fc, = allowable working unit stress, lb. per sq. in., for long 
columns. 


49 Mitteilungen wber-Versuche ausgefiihrt vom Ejisenbeton-Ausschuss des 
ésterreichischen Ingenieur-und Architekten-Vereins, ‘‘ Versuche mit Eisenbeton- 
saulen,” Heft 3. 

50 Knickungsversuche mit Eisenbetonsaulen Zeitschrift des Vereins Deutscher 
Ingenieure, Prof. C. Bach, 1913, p. 1969. 


94 TESTS OF REINFORCED CONCRETE 


Where f,= allowable working unit stress, Ib. per sq. in., for 
short columns; 
A = cross section of column, sq. 1n; 
1 = length of column in feet; 
I = moment of inertia inch units of the cross section of 
the column. 


TORSIONAL RESISTANCE OF CONCRETE AND REINFORCED CONCRETE 


Character of Stresses Due to Torsion.—Torsion is caused by 
moments of opposite signs applied at ends of a member and acting 
in planes normal to its axis. This produces a tendency of two 
adjoining normal planes to slide upon each other, which results in 
shearing stresses in the normal planes. The magnitude of the 
shearing stresses is zero in the centroid and increases according to a 
straight line to a maximum at the periphery. In members with 
circular sections the maximum shearing stress is uniform over the 
circumference of the section. In square and rectangular cross sec- 
tions, the shearing stresses along the. periphery are not uniformly 
distributed, but vary from a maximum in the center of the sides to 
zero in the corners. In rectangular sections the shearing stress in 
the middle of the long sides is larger than in short sides where the 
failure of rectangular sections starts. 

An additional effect of torsion is elongation of the sides and par- 
ticularly of the edges which produces tension. In concrete speci- 
mens, the tension takes place along the surface at approximately 
45° to the axis. | 

This tension (and not the larger shearing stresses) is the immediate 
cause of failure because concrete is weaker in tension than in 
shear. There are no formulas for computing the tensile stresses, 
therefore the torsional stresses as given below may be used as their 
measure. | 

Ultimate Torsional Stresses.—The ultimate torsional stresses in 
homogeneous beams may be computed from the following formulas: 


Let /, = ultimate torsional unit stress in concrete, lb. per sq. in; 
M, = the torsional moment in inch-pounds; 
b = the short side of the section in inches; 
h = the long side of the section in inches; 


d = diameter of circular section. 


TORSIONAL RESISTANCE OF CONCRETE 95 


For rectangular and square sections, 


PE noel ed 0 
0.45 + 7 





For circular sections 


16 M, 
ie an TREES ° 2 . . . . ° ’ eri 3) 


Bach’s tests,°' made by C. Bach and O. Graf in Stuttgart, Ger- 
many, to determine the resistance of concrete and reinforced con- 
crete to torsion, consisted of two groups of specimens: (1) plain con- 
crete with square, rectangular, circular, and circular ring cross 
sections respectively; and (2) square and rectangular reinforced 
concrete with varied amounts and dispositions of reinforcement. 
The length of all specimens was 6.4 ft. and the cross section was 
11.8 in. square, 8.3 by 16.6 in. rectangular, and 15.8 in. diameter 
for circular specimens. 

Method of Testing.—The specimens were tested by applying 
torsional moments at the ends. An initial moment of 21 670 in.-lb. 
was applied first and the instrument read. The moment was 
then increased, by successive increments of 21670 in.-lb., until 
failure. 

Torsional Resistance of Plain Concrete.—The failure occurred 
in the center of the specimens by cracking at 45°. Failure always 
followed closely the appearance of the first crack. In specimens 
with rectangular cross sections, the first crack started on the wide 
face. 

The ultimate torsional unit stresses, figured by the formulas for 
homogeneous beams, are given in the table on p. 96. 


51 “ Versuche iiber die Widerstandsfachigkeit von Beton and Eisenbeton gegen 
Verdrehung,”’ by C. Bach and O. Graf, Berlin, 1912, Heft 16. 


96 TESTS OF REINFORCED CONCRETE 


Ultimate Torsional Unit Stresses. (See p. 96.) 


Concrete, 1: 2:3 by volume. Aggregates, Rhine sand from 0 to 4-in. 
diameter and Rhine gravel from }-in. to j-in. Average compressive strength of 


12-in. cubes at 45 days, 3 540 Ib. per sq. in. Age of specimens at test, 45 days. 


Compiled from Tests by C. BacH and O. GRAF 





Ultimate Torsional Unit Stress 


| 
| 
| 
Cross Section | In terms of In terms of 








Lb. per sq. in. tensile compressive 
strength strength 
Squarehces. ean ee ee | 432 1.62 0.12 
Rectangularyes. oa) Sees 462 1/78 0.13 
Cireulat jos ad ese een eee ! 364 1.38 0.10 
Circular, Rings). as24ese eee 243 | 0.92 0.07 


Oa 


Torsional Resistance of Reinforced Concrete Specimens.—Longi- 
tudinal reinforcement had very small influence on torsional resistance. 
For specimens reinforced with 1.13 per cent and 2.26 per cent of 
straight bars, the increase of torsional resistance was only 9 per cent 
and 14 per cent, respectively. More marked was the influence of 
inclined bars. For specimens with 1.13 per cent of bars at 12°, 
the ultimate resistance was increased by 27 per cent over that of 
plain concrete specimens. 

Stirrups were much more effective in resisting twisting. For 
specimens with 2.26 per cent of longitudinal bars and 0.267 in. 
diameter, stirrups spaced 3.93 in. on centers, the torsional moment 
at first crack was 31 per cent larger and at failure 66 per cent larger 
than for plain concrete specimens. 

The most effective reinforcement consisted of stirrups the legs of 
which were in planes parallel to the sides, but were inclined at 45° 
to the axis of the specimen. Evidently, the stirrups resisted directly 
the tension acting at 45° to the axis. Specimens reinforced with 
2.26 per cent longitudinal bars and stirrups 0.276 in. in diameter, 
spaced 3.7 in. on centers inclined at 45°, showed an increase in 
resistance of 55 per cent at first crack and 134 per cent at failure over 
plain specimens. 


TORSIONAL RESISTANCE OF CONCRETE 97 


Toronto Tests.—Tests made at the University of Toronto by 
C. R. Young, W. L. Sagar, and C. A. Hughes °2 gave the following 
results: 


Ultimate Torsional Unit Stresses—Toronto Tests 


Concrete, 1:2:4. Strength of 6 by 12 cylinders at 28 days, 1700 lb. per 
Sq.in. Age of specimens at test, 28 days. 


Compiled from tests by C. R. Youne, W. L. Sacar, and C. A. Hucuzs 


| | 





Ratio to 
Ultimate Average 
Specimen Actual Longitudinal Spiral Rersken torsional 
Cross , : Unit Strength 
Number é Reinforcement Reinforcement ; 
Section Stress of Plain 
lb. persq.in.] Concrete 
Specimens 
Al 4.985 None None 539 
A2 DO X75 faa o: 467 
A3 5.0 10.09 Ly oe 602 
Average 536 
Bl 5.06X5.0 |4rods 1 in. diameter ‘ee 532 0.99 
2) |5.06 7.59 | 4 ** J ‘‘ * Sa 506 0.95 
B3 DOO, LOCO Oe ges ae es 584 1.09 


Average 541 | Average 1.01 


C1 5.138X5.19 |4 ‘ 


al 
« 
. 
- 
- 


4—5 deg. spirals of 


75-in. iron wire 668 1.25 
Oops Weg Be OS ie aa yy 621 1.16 
C3 BIS M1031 45%) 4 r sf 530 0.99 
Average 645 | Average 1.21 
Omitting C3 | Omitting C3 

D1 SOK Oseh a fhe Rt a 8-45 deg. spirals of 
#-in. iron wire 838 TG 
D2 |5.19X7.81 | 4 a, be Ss he 740 1.38 
D3 9.13% 10.38)4 “* 3 * he vid 790 1.47 


— —_—— 


Average 789 | Average 1.47 

er ere er ET ee Be eee Pa 
If the excess of torsional moment over that for corresponding 
plain concrete specimens is assumed to be resisted by the steel 


* Torsional Strength of Rectangular Sections of Concrete, Plain and Rein- 
forced, by C. R. Young, W. L. Sagar and C. A. Hughes. University of Toronto 
Bulletin No. 3, 1922. 


98 TESTS OF REINFORCED CONCRETE 


spirals, the computed stresses in reinforcement will be as given in 
table below. 


Stresses in Spiraling at Failure of Specimens—Toronto Test 


ee 


Excess of Torsional Moment over that 





sas for Corresponding Plain Concrete Pikes Sa 
Number ae lb. per sq. in. 
pecimen, in.-lb. 
i Se 

Cl 5 100 18 000 

C2 8 800 31 000 

C3 Unreliable ~ ——s "sae Re 

D1 11 100 19 500 

D2 15 800 27 900 

D3 16 000 28 200 


oe ee ee 


In the above computations, the excess of torsional moment is 
assumed to be resisted by tension in the sprial, cut by a normal plane 
and equal part by compression in concrete acting at right angles to 
the direction of the reinforcement. 


TESTS OF FLAT SLAB CONSTRUCTION 


Numerous tests have been made of flat slab construction, most of 
which, however, were on completed buildings, where of necessity the 
tests were restricted to loadings well within the elastic limit of the 
materials. While the results of these tests are of interest, as demon- 
strating the strength of the construction, they may be misleading when 
an attempt is made to estimate from the stresses in steel the ultimate. 
strength of the slab. The measured steel stresses in all these tests 
were comparatively low. This fact has caused considerable misun- 
derstanding. The final strength of the slab has been often based 
upon this low steel, with the consequence that the strength so esti- 
mated was greatly in excess of the actual strength. 

The low steel stresses found in tests on completed buildings are 
due to the fact that the slab is still in the stage where a large propor- 
tion of the tensile stresses is resisted by the concrete. A small increase 
in loading is often sufficient to overcome the tensile strength of con- 
erete, when the bulk of the tensile stresses is thrown upon the steel 
and a considerable jump in tensile stresses takes plarz To eliminate 


TESTS OF FLAT SLAB—TWO-WAY SYSTEM 99 


the uncertainty, it is necessary to carry the test of the slabs to a point 
where the concrete cracks sufficiently to transmit all the tensile 
stresses to the steel. The two tests described in the following pages 
were conducted practically to destruction on specially built panels. 
They are much more instructive than the tests conducted on com- 
plete buildings. Both of the tests were made by Professor W. K. 
Hatt at Purdue University, one for the Corrugated Bar Company 
and the other for Mr. Edward Smulski. 

In both tests stresses in steel and concrete were measured as 
described on p. 113 in connection with tests of reinforced concrete 
buildings. 


TESTS OF FLAT SLAB—TWO-WAY SYSTEM 


Professor Hatt conducted several tests for the Corrugated Bar 
Company, one of which is reported in the Proceedings of the American 
Concrete Institute, 1918, p. 177. The information given below is 
taken from this report. 

The test slab consisted of four panels 16 ft. square with 6 ft. 
overhang on three sides and a wall beam on the fourth side. The 
object of the overhang was to balance the load at the column and 
produce in the adjoining panel the same conditions as in interior 
panels in a building. The 6 ft. proved to be greater than necessary 
for the purpose. A 4 ft. 6 in. overhang used in an earlier test for 
Mr. Smulski, described later, proved too small. The dimensions 
are shown in Fig. 34. 

The reinforcement is shown in Fig. 35. The steel was disposed 
according to the arrangement of the Corr-Plate floor, as designed 
by the Corrugated Bar Company, and the amount of steel was that 
necessary to meet the requirements of the moment coefficients 
recommended in the Joint Committee report of 1916. (These 
coefficients are considerably larger than recommended by Joint 
Committee report of 1924.) 

The design load was 150 lb. per sq. ft. 

The steel was hard grade steel with a yield point of 55 600 Ib. 
per sq. in. The concrete had a compression strength of 2 300 lb. 
per sq. in. at the age of twenty-eight days, and 2 586 lb. per sq. in. 
at ninety-five days. 

The slab was tested at the age of sixty-six days. 


100 TESTS OF REINFORCED CONCRETE 


Loading.—The slab was loaded with brick. Loads of 150 lb., 
300 lb., 450 Ib., 600 lb., and 803 Ib. per sq. ft. were applied suc- 
cessively. 

Steel Stresses were measured with the 10-in. Berry Extensometer. 
The results are given in Table on p. 102. 


35 


{cae Seem eS SS SORE SES ANTS 


nee ee eS Se SE = -- 38' 





Section A-A 
Fic. 34.—Concrete Dimensions, Slab J. (See p. 99.) 


Maximum Load.—The maximum uniform load was 595 lb. per 
sq. ft. over the whole slab. After this was entirely removed, the 
brick were piled to a load of 803 lb. per sq. ft. over one panel, the 
overhang, and parts of the adjoining panels. 

When the load of 600 lb. was taken off the slab, the large cracks 
closed, as they would be expected to do, since the steel had not 


reached its yield point and there was no compression failure in the 
concrete. 


TESTS OF FLAT SLAB—TWO-WAY SYSTEM 101 


The effect of the load of 803 lb. was to dish down the panel and 
crush the slab at the column heads. The failure of the slab at these 
column caps began with a feathering of the concrete in the drop at 
the edge of the column cap, followed by a diagonal failure in the drop 
around the head of the column. The punching shear at this stage 
was computed to be 198 lb. per sq. in. 




































































North 
fasta pallu pa dae 3a gene : ABE agticg 
any SLABS BOSE RH RVG 9 ss SOR NH SSS ee 
wt S Sat > 203" my SAS eas eZee 976 SoS > weary O 
1 2 Wom 213 3-" 2529" 
{ 1 
iy 71 Ns. 4'%721.8% 3 
HEE es 
4 ! 
‘dae 3%" 98's 
i ! 2-7 25+9 
Fe: 
ee tn 
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' 1 
os 
1 a 
2 
ibe 
1 { 
' { 
ee ee alse oe 2-34'°928"5 
; ' a 4,"0 
Sa PS ee Nest |5- "92.8% yy 
@ at ll9; o AS 5-" 2549" 3 
+ r H 
soe tye ee oy 2.8 ae roc ry 3-3"° 9-8" = 
a bel date tee 
Fier GA] 5-34" 72" re 
ae 5-9 2753" 
We ee 
1 Sy 
a ow 
ia i 
' ! 
( ' 
tis’ | 
at ——3|3-76°"/2"8"s 2-99"°9-8"s 
' 2-” 21-3"b 3-9) 25UQM 
fe by | 
t TH (% 7794 a" Ure A 
wy Wore ; 5-2  9°8"s 
4 a eee 263") 5" 25°9"% 
Iixie 10-129 4-6"6 TB. ays 
‘ oN 2-3" °272-8"s 6-72 ©916"6I, 7 wo Ke) 
jo tangas [atease twas 
ae TNT "7 122 8"3 SSN SH T- BS SAS 
Viv |) Sate 223 bSadead Odo eel % ine 
3 hewn: 3 (lieeos es 
Sonn nnanan- a a ee 186 
Seger tage ag 2s enn Ro ROSH ae agen ene eli hy HG, A 


South 
Fig. 35.—Steel Layout, Slab J. (See p. 99.) 


At load of 600 Ib. per sq. ft. the stresses were: 

Average stress in band that was stressed to the greatest degree, 
56 000 lb. per sq. in. 

Maximum compressive stress, 1 800 lb. per sq. in. 

Ratio of yield point of steel to stress, 1.40. 

With poorer concrete, the compressive strength of the concrete 
and the yield point of the steel would have been reached earlier. 


102 


TESTS OF REINFORCED CONCRETE 


Steel Stress and Moment Coefficients in Slab J—Interior Panel—Live Load 


Band 


Average.... 


Average... « 


Average.... 


A |Positive N-S.. 
Positive E-W. 


Positive N-S.. 
Positive E-W. 


Positive N-S.. 
Positive E-W. 


m = moment coefficient 


150 


Stress 


1 470 
3 110 


2 290 





1 340 
2 510 


1 925 


1970 
1 890 








1 930 






































lb. 300 lb. 450 lb. | 600 Ib. 
m Stress m Stress | ~™ Stress | m 
5 820 15 580 28 780 
9 620 16 040 30 650 
0.0100| 7 720/0.0166; 15 810/0.0225 29 71510.0315 
6 000 16 450 32 940 
10 590 20 490 36 740 
(0.0048! 8 29510.0106| 18 470 0.0154| 34 840)0.0220 
5 600 15 650 31 820 
10 190 18 870 33 420 
0.0024! 7 895)0.0048)| 17 260/0.0070 om 620,0.0100 
0.0130 32 300|0.0184 











| | 
Positive, Whole Av. | 2 048|0.0048 7 970|0.0095| 17 180 




















35 040 
33 020 





45 740 
55 810 





50 028/0.0480 


| 56 290 
55 500 





A |\Negative N-S.| 2 660 8 617 20 880 
Negative E-W| 3 590 10 495 21 778 
| Average....| 3 125)0.0228) 9 556 0.0349] 21 328/0.0515} 34 030\0.0655 
B \Negative N-S.| 2375 8 275 24 320 
Negative E-W| 3 680 11 450 29 675 
Average....| 3028 9 863 26 748 
C |Negative N-S.| 2170 TAGS 33 890 
Negative E-W| 3 110 8 010 29 370 
Average....| 2640/0.0027| 7 590 0.0041) 31 630/0.0114 





| | | 
Negative, Whole Av.| 2 931 0.0100; 9 000/0.0151) 26 570 








55 895)0. 0150 


0.0270) 46 650/0.0355 








Pntires. 2) 22 








2 478 0.0148) 8 480/0.0246) 21 870 











(0.0400) 39 500 0.0539 


TESTS OF FLAT SLAB—TWO-WAY SYSTEM 103 


Distribution of Stress —The ratio of negative to positive stress is 
as follows: 


Band Load 450 Load 600 
Ee ert ee a. ee owak sla. Lo 1.14 
ee og che ky bs rr 1.45 1.43 
eM ee es ok haw na 1.86 1.72 


Evidently, the negative steel may well be increased, especially in 
the margins or mid-section. 

The uniformity of stress is shown in the following, where the 
average = 100. 

Range of minimum to maximum band stress: 


Load 450 Load 600 
ORES Se fo OE a Se 92-107 92-107 
VASO. tees Duce, Tensei an a i aa 82-120 73-120 


Deflections as a Fraction of the Direct Span (16 ft.)— 


Load | Direct Span Diagonal Span 
pote oms ere a). te Mie cas, 1/3800 1/1920 
Fem Bo Fdieaps cys goo «GS ave os 1/1370 1/735 
OE RS ES OS be ae 1/580 1/315 
Fy me ee Ne gee ek s 1/375 1/194 


Distribution of Moment Coefficients—The total coefficient is 
divided in per cent. 





Slab J 

Positive 450 lb. 600 Ib. 

Gh . 43.0 43.2 

| See oot 29.6 

beatin. 3 BILE we 

ELD) s2baselO0. 0-99 34,0 

Negative 

nor: ack |. 3 645.2 

Tee 3109 34.1 

CR ie 


. 20.6 20.7 


1002085 67-7 100.0%" (66.0 


Load Carried by Lintel Beam.—'The steel stresses at the center 
and at the support of the lintel beam allow computation of the total 
moment of resistance of this beam. This may be equated to $W1, 
and the value of W, the load causing the flexure of the beam, may 
be computed. In Slab J, at 600 lb. per sq. ft. loaded on the panel, 
the indicated load on the lintel beam was 0.121 of the total panel 


104 TESTS OF REINFORCED CONCRETE 


load. Another slab, with more highly developed steel stresses, 
yielded a fraction of 0.16. 
The moment in the wail column was 0.02 WI where W=load 
on panel and /=panel length. 
Comparison of Wall and Interior Panel.—The ratio of positive 
stress in wall and positive stress in interior panel was: 








Loading 
450 lb. 600 lb. 
Acoycia bias See NS Fak OE ae ee ee OFS 0.73 
Bw begins ath Sead Cen ek ean ae 0.91 0.82 
Ca ene MMR PR UN in Ae rae ey 1.01 0.93 
Average.) Jueed. «os en Fee ae ee 0.89 0.83 


The 20 per cent additional steel in the wall panel positive sections 
is evidently sufficient. 

The stresses in the negative steel running into the wall beam, 
2 680 Ib. per sq. in., and into the head of the wall column 10 230, 
are, at 600 lb. load, low compared to the negative stresses in the 
interior panel. Of course, this is only an expression of the yielding 
of the wall beam and wall column. 

Measurements on the web of the lintel beam gave consistent 
indications of torsion. 

The distribution of compressive stresses over the drop panel 
shows approximately that the concrete compressive stress at the 
edge of the drop is only about 25 per cent of the stress at the edge 
of the head of the column. In considering the allowable unit stress 
computed as an average over the drop panel, the maximum that 
fixes the strength of the slab must be considered. 

Various diagrams are reproduced, showing distribution of steel 
and concrete stresses and photographs of the slabs. 


TEST OF FLAT SLAB—SMULSKI SYSTEM 


The test was made by Professor Hatt for Mr. Edward Smulski 
and was reported in the Proceedings of the American Concrete Institute, 
1918, p. 206. The information given below is taken from this 
report. The test slab consisted of four panels 16 it. square with 
overhang on three sides and a wall beam on the fourth side. The 
dimensions of the panel are shown in Fig. 36, p. 105. 

The overhang was intended to balance the bending moment at 
the column and thereby reproduce in the panels the conditions that 


TEST OF FLAT SLAB—SMULSKI SYSTEM | 105 


exist in interior panels. Test showed that the 4 ft. 6 in. overhang 
was not sufficient. In a subsequent test made by the Corrugated 
Bar Co., and described on p. 99, an overhang of 6 ft. was used. 

_ The reinforcement is shown in Fig. 37. For fuller description 
of the system see p. 367. 


yeaa sci ipte ate oe SB0 Oey s eh aec sd ele ea ores ee tees: ie) ee a rie 
a! 


© ae ee 
a 








Section A-A 
Fic. 36.—General Dimensions of Smulski Test Slab. (See p. 104.) 


The design load was 150 Ib. per sq. in. and the amount of steel 
was that necessary to meet the requirements of the moment coeffi- 
cients recommended in the Joint Committee report, 1916. The 
two panels adjoining the overhang were treated as interior panels. 

Steel of structural grade was ordered. From test coupons it was 
found that a large part of the steel was of intermediate grade with 


106 TESTS OF REINFORCED CONCRETE 


elastic limit of 42,200 lb. Slab was built Nov. 30, 1916. The test 


began Jan. 29, 1917 and extended over a period of forty-six days. 


Loading —The slab was loaded with-brick. The load was 


applied in the following increments: 


Load, lb. per sq. ft. 
East panel....... 150 300 450 450 3875 475 475 
West panel....... 150°” 300 450. 625 700— 700 to, 950 70 




















Bey 





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: North Haff of Unit Ce same as Unit C 


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Sr Pk 






Sree ey, 3 
waa Seraeeea a ca 


Spandrel Beam 9"x16" 





—Radials _---Center Rings 


Reh Mame earn 


‘Truss Bars 


Section Thru Colurnn Head 


Fic. 37.—Arrangement of Reinforcement in Test Slab. (See p. 105.) 


The loads in the fourth to seventh increments were unbalanced and 
were much more severe on the two west panels than would have been 
the case with uniform loading. 


TEST OF FLAT SLAB—SMULSKI SYSTEM 107 


Steel stresses were measured with 10-in. Berry Extensometer. 
The results are given in the tables on p. 107 and 108. 


Summary of Steel Stresses, Top of Slab, in Lb. per Sq. In. 
(Zero at 150 Lb. per Sq. Ft.) 





Columns 





Position of Gage Lines 


Loading of Panels, lb. per sq. ft. 











Quadrant Between|West 300; 450 625 700 | 700 
rings |East 300} 450 450 | 950 | 
475 | 700 
Radials, Unit C 

Center column. . |Northwest... 1-2 3 360 5 760} 8 980] 33 000) 42 800 
2-3 2 460 3 060} 7 180} 26 400) 42 700 

3-4 PAG Tee eg eee ee 13 200} 35 700 

4-5 BELO. Pate et cls s 1h Oe se 16 800 

Southwest... 1-2 SABO GRY o.: 8 080) 34 200] 41 100 

2-3 2640 | 2280] 7 240] 26 400] 31 500 

3-4 1 440 900} 3 220] 14 400} 21 900 

4-5 L260 eee 2980) fos 12 600 

North column.../Southwest. . . 1-2 4860 | 5040} 10 880] 39 600/ 42 600 
2-3 5 580 | 6900} 9 700) 22 200] 48 300 

3-4 3 960 | 2400) 7 360} 24 000] 42 600 

4-5 4240 | 3420] 7 780/19 800} 21 600 

Southeast... . 122 7 260 | 8580} 6 520/17 400} 44 700 

2-3 7800 | 7680} 6 460) 19 200] 40 800 

3-4 4620 |11 340} 6 700) 19 800/31 800 

Be sa cent 2 5 400} 9 280} 12 000! 36 000 

Rings, Unit C 

Center column. . |Northwest... 2-5 D 460724. 54 es 8 110} 24 600; 40 600 
Southwest... TOGO Tes es 8 040} 27 320) 38 400 

Southeast... . 4900 | 5280} 5 920) 16 140} 33 300 

Northeast... . 5700 | 4800] 6 640; 21 500) 41 940 

North column...|Southwest. . 5 3800 | 6120) 8 980} 18 780) 34 000 
Southeast . 4380 | 5600! 5 380/13 200] 34 000 

South column... |Northwest... OOO wate 7 560} 14 340) 28 700 


108 TESTS OF REINFORCED CONCRETE 


Summary of Stresses, Bottom Reinforcement, in Lb. per Sq. In. 
(Zero at 150 Lb. per Sq. Ft.) 


Position of Gage Lines Loading of Panels, Lb. per Sa. ft. 


Unit West 300 | 450 | 625 | 700 | 700 
Ring Number Quadrant or Panel 950 
East 300 | 450 | 450 475 700 


Deena ee Pee aa Re a ne i 


Unit A ANGOvA ei aa oe ae Northwest....-<5- 3 240] 6 O60!..... 
Southwest......--- 3 000) 5 340|..... 
Southeast.....>. 












14 040) 49 800 
43 680) 49 800 
12 580) 23 640 


Unit, Aw.h opie cas tee Sia: “ance West panel........ 26 240| 34 620 
Hast panel. vs ws. - PUTA beni Pete cts ics cil ids oo 18 840 





( (East panels) | Wall panel........ 3 480/001 thie 2 eee 22 500 











Units B | 5456 Intermediate panel. 5 B80) Jas Teak eae 34 740 
(West panels) | Wall PAUEl dy e-Mat 3 960] 6 360| 7 420] 30 540) 36 720 

Intermediate panel. 3 900| 5 400] 6 070} 42 120; 49 800 

(i(ast). eee2h Wall panel........ 2 S20! <a ch asus Cleans 10 440 

eae. Intermediate panel. 3. 420) Reraee te teva teen ees 30 000 

CW est) eles Wall panel........ 1 680| 2 640] 6 220] 14 280} 16 450 


7 180 


ar 





Intermediate panel. 





26 jit 36 360 


Sa umencnrn nena mee 


Note that at final load the west panels, which were subjected to the heavy unbalanced 
load, show much higher stresses than the east panels. This is due to yielding of columns 
under unbalanced load. 


Analysis of Steel Stresses. 


Design Load of 150 lb. per Sq. Ft.—The steel deformations for 
this load were often so small as not to be apparent on the Berry 
Extensometer, in which one division on the dial indicated about 
600 lb. per sq. in. In the rings, in a number of cases, small com- 
pression stresses were observed at low loads. These stresses, which 
were probably due to uncorrected temperature effects, later changed 
to tensile stresses. The tensile stresses, when indicated, were 1 000 
to 3.000 lb. per sq. in. The stresses ‘are therefore quoted in tables 
as from zero at 150 lb. per sq. in. 

At Time of Yielding of Columns.—At a load of about 625 lb. per 
sq. ft. on west panels and 450 Ib. on two east panels, the outside 
columns of the more heavily loaded panels began to yield, and cracks 
began to open on the under side of column heads and brackets. T his 
load may be considered as the end of the first phase. At this time 


TEST OF FLAT SLAB—SMULSKI SYSTEM 109 


there were none but fine cracks in the slab and the measured stresses 
were as follows: 

Stresses at 625 Ib. on West Panels and 450 Ib. on East Panels. 
(Zero at 150 lb. per sq. ft. load.) 


Unit B. Bottom of Slab. 


Rings 1-2................. 6180 interior panel 6220 wall panel 
re OTL Oe a 6680 Ne 2 FAZU ES ES 
PRUs TOGS oye Gk ee 7000 oy FUGUE 
a mee ay 7230 all panels 
Pheer rt oe ek yc 8440 inner gage lines 6680 outer gage lines 


The above stresses were observed in the more heavily loaded 
panels. 

Stresses at Greatest Applied Loads.—At the final readings of steel 
deformations, the slab was in the second phase. 

In analyzing the stresses at higher loads, it must be remembered 
that after the load exceeded 625 lb. per sq. ft. because of cracking of 
columns the slab entered the second phase, in which, by gradual 
opening of the cracks under the brackets and column heads of the 
outside columns, the restraint of the slab was gradually removed. 
This caused gradual increase in bending moments, both in the center 
of the slab and at the column, and explained why, under maximum 
load, the stresses-and deflections increased gradually until they 
reached their maximum after twelve days. 

The stresses at the maximum applied load include, besides the 
stresses due to bending moments, (a) stresses caused by plastic 
deformations under the heavy load standing on the slab for a long 
time; (b) stresses caused by the change from a slab restrained at 
wall columns and continuous in all other directions (as contemplated 
in design) to a slab nearly freely supported at the outside columns. 

The maximum measured stresses in Ib, per sq. in. were as follows 
(Zero at 150 lb. per sq. in.): 


ete a ee, 49 800 west panel 23 640 east panel 
Unit B—Rings 1-2......... 33 180 interior panel 13 490 wall panel 
SPSS aes. 42840 ‘ sé 2 GORE SOP NSS 
Umit C—Rings........... .. 35800 
veces 2). sete, 845,” 42 840 inner gage lines to 21 750 outer gage lines 


BAIR, ete eT atti: 46 250 


110 TESTS OF REINFORCED CONCRETE 


Points of Inflection—The distance of the points of inflection 
are given in the table below. 


Rectangular direction.........- Span = 16 ft. 0 in. Clear span A i rolers Sao 
Diagonal direction...........-- Span = 22 ft.6in. Clear span = 19 ft. O in. 


l| 


ee asic eee ae 









Distance of Points of Inflection 


LO ee eae 











From Center of From Edge of 
Position Direction Column column head 
Ft. I In terms of Rial Tae In terms of 
span clear span 
Center column... .s-.. Northeesss BAT 10 0.24 Pa ee 0.15 
Northwest 5 2, 0.23 3 2 0.16 
SHuthee tee 4 2 0.21 2 4 0.20 
Southwest 5 8 0.26 3 | 10 0.20 
West genter eolumn.. .-\Hast.. 4 - a5 4 | 10 0.30 4 0 0.24* 
Northeast 5 2 O25 sf 5 0.18 
Northwest column... ..|Southeast 5 0 0.22 3 c 0.17 
Center wall column....|North.. 4 1 0.26 2 0 0.17 
Northwest 5 6 0.25 4 2 0-22 





Southeast wall column. |Northwest... 


ee 


* Doubtful. 


Distribution of Stresses in Rings.—The distribution of stresses in 
rings is evident from Fig. 38, p. 111, and Fig. 39, p. 112. The 
figures were drawn by plotting in the center of each gage line the 
observed stresses. 

Comparison of the Two Flat Slab Tests.—The two tests described 
above were made by the same experimenter and the general dimen- 
sions of the test specimens were practically the same. The results 
are, therefore, comparable. 

Before making comparison, a proper allowance should be made 
for following differences in design and testing. 

(1) The projections in the two-way system slab were larger than 
in the Smulski System. The larger projection increased the restraint 


TEST OF FLAT SLAB—SMULSKI SYSTEM 111 


of the slab at the columns thereby reducing the stresses in the positive 
reinforcement of the adjoining panels. 

(2) The wall columns were evidently made stronger in the Two- 
way System test because the cracking below the bracket, observed 








> 


Col.C2 


| 
oh Ae |. 

if 

ys 


Col A2 


eeeUPAe Re) MA Col BI. a yh iN) osu live lates 


Sj 
>: 


Y 





Unit Ce-Ring 
No.6 


Fig. 38.—Distribution of Stress in Rings of Column Head Unit C. (See p. 110.) 


in earlier test, did not take place. Therefore, the results of the two 
tests are comparable only for the loads prior to those producing 
cracking of wall columns in the Smulski System. 

(3) As evident from compressive tests of cylinder, the concrete 


112 TESTS OF REINFORCED CONCRETE 


in the Two-way System was of better quality, presumably because 
+t was cured under more advantageous conditions. (Smulski slab 
was built in winter while the Two-way slab was built in the summer.) 

(4) The distribution of test load as adapted by Prof. Hatt for the 
two tests differs to some extent. 





Fic. 39.—Distribution of Stress in Rings of Unit Be. (See p. 110.) 


In the earlier (Smulski) test, open spaces were left at two columns 
for observation purposes, and extra brick was placed in the central 
portion of the slab to compensate for the omitted load and to pro- 
duce same bending moment as would have been produced by a full 
uniform load. 

In the second (Two-way) test practically the whole area was 
loaded and observations were made in tunnels. 

Due to this difference in loading, for equal unit load in the two 


TESTS OF REINFORCED CONCRETE BUILDINGS 113 


tests, the Two-way slab was subjected to larger external shears, 
although the difference in bending moments was not appreciable. 
Keeping these differences in mind, the stresses as given in table 
on p. 102 may be compared with the stresses for equal loading as 
given in table on pp. 107 and 108. 
The stresses for the Smulski test are given more in detail while 
for the Two-way System as averages for whole bands. 


TESTS OF REINFORCED CONCRETE BUILDINGS UNDER LOAD 


One of the most important developments in testing within the 
last few years is the testing of complete structures under load. It 
has been customary to make deflection tests of engineering structures 
to determine whether the structure can safely carry the load for 
which it has been built. Such tests, however, are of little scientific 
use as they do not give the stresses in the structure. Sometimes 
they are actually misleading, because with small deflection there 
may exist in certain parts of the structure stresses that are much 
higher than allowable. Moreover, such tests do not determine 
_ weakness of details or the effect of continuity. 

The recent tests on completed structures are much superior to the 
old deflection tests, as they measure not only deflection of the 
structure, but also the stresses in various parts of the members. 
These tests were inaugurated by Prof. Arthur N. Talbot of the 
University of Illinois, with the assistance of Messrs. A. R. Lord and | 
W. A. Slater. The first building tested in this way was the Deere 
and Webber building in Minneapolis, in October and November, 
1910. Following this test, several other tests were made under the 
auspices of the Reinforced Concrete Committee of the American 
Concrete Institute. 

The instruments used in such tests are (1) extensometer for 
measuring the stretch or compression of the materials, (2) deflecto- 
meter, for measuring deflection. 

The extensometer consists of a framework (which in the best 
instruments is made of invar steel to prevent appreciable changes in 
length due to the changes of temperature), two movable legs attached 
to the framework and provided with sharp points, and a device for 
measuring accurately any changes in distance between the points. 
In order to find the stretch in steel, the bar to be tested is uncovered 
in two places a few inches apart, and small holes, called gage holes 


114 TESTS OF REINFORCED CONCRETE 


(0.055 in. in diameter) are drilled. An observation on the gage line 
is taken before the structure is loaded and again at each increment 
of the load. The difference between the original reading and the 
reading at any load gives the stretch of the steel due to that load. 
The stress is then found from the known relation between the defor- 
mation and the stress. 

The compression in concrete is measured by making small holes in 
the concrete, inserting metal plugs, and then marking the gage holes 
in these plugs as was done for the steel. Readings and stresses are 
obtained in the same way as for steel. Since concrete flows grad- 
ually under heavy loads, the readings must be made immediately 
after each loading. 

For measuring deflections, a rigid scaffold is built directly under 
the members to be tested. In the place in which a deflection reading 
is desired, a steel plate is fastened by plaster of Paris to the under 
side of the beam or slab. On a vertical line below this steel plate, a 
steel rod is fastened to the scaffold. Before the test is begun and at 
different stages of loading, the deflectometer is placed between the 
plate and the rod below, readings are taken, and the difference 
between the original reading and the readings under the load give the 
deflection. 

For loading the panels, there may be used: (a) brick, (0) cement in 
sacks, (c) loose sand in boxes or in sacks, and (d) pig iron. In making 
tests, care always must be taken that the material does not arch. 
The whole floor cannot be covered with the load, because places must 
be left uncovered for the taking of measurements, and aisles must be 
left to make the points accessible. 

It is important that the test load should cover a sufficient floor 
space to insure that certain parts of the floor resist nearly the full 
load which, in the calculations, they are considered to take. 

Wenalden Building Test.®*—The floor panels in this building are 
15 ft. by 20 ft. The slab is 33 in. thick; the girders, placed between 
columns in the short direction, 74 in. by 20% in., reinforced with 
four 3-in. square bars in the middle. The longitudinal beams, 5 ft., 
apart on centers, are 6} in. by 18% in., reinforced with four {-in. 
square bars in the middle. Half as much steel was used over the sup- 
ports as in the center. 

The floor was designed for a live load of 200 lb. per sq. ft., and the 
total test load was made 400 lb: per sq. ft. and placed in layers of 

63 University of Illinois Bulletin No. 64, January 13, 1913. 


TESTS OF REINFORCED CONCRETE BUILDINGS 115 


80 Ib. per sq. ft. A set of observations was taken after every addi- 
tional loading. 

The measurements were taken in steel at the support and at the 
center of the span; also measurements of stresses in concrete at the 
support and at the center of the span. 

This test proved conclusively that the beams and the girders act as 
continuous ones. While the stresses in steel in the center and at the 
support were not excessive, the highest stress being 17 000 lb. for 
the total test load, the stresses in concrete at the supports of the 
_ beams were high and in some places even reached a stress of 2 200 lb. 
per sq. in. Even under the working load, stress in concrete was 
1 150 lb. per sq. in. The compressive stresses in the center of the 
beam were low, and it appeared from the test that the total slab 
acted as a compressive flange of the T-beams. It must be noted 
that in this case the overhang of the flange was seven times the thick- 
ness of the slab, while in practical design, we consider only an over- 
hang of six times the thickness of the slab as effective in taking 
compression. The total compression in a beam, figured with the 
assumption of a straight line distribution of stress and no tension in 
concrete, was much larger than the total tension, showing that either 
arch action existed in the beam or considerable tension was carried 
by concrete. ‘The difference was especially large at the supports, 
where the tension must have distributed itself over the entire slab. 

Test Cracks.—Tensile cracks were observed in the middle portion 
of the bottom of the beams. They formed at the same stress in steel 
as is usually found in the laboratory. Diagonal cracks developed in 
the girder which carried a very large shear (V = 40 000 lb. and v = 
360 lb.), just outside the junction with intermediate beams. The 
cracks were inclined at about 45°. They did not close entirely after 
removal of the load. It is supposed that the restraint at the ends 
prevented fuller development of the cracks. (See also p. 66 on 
formation of cracks in continuous T-beams.) 

Deflections.—The deflections offered further proof of the contin- 
uity of the beams, being much larger, in the middle panel, for one 
panel loaded than for three panels loaded, as would be expected 
from a continuous beam. With three spans loaded, deflection of 
intermediate beam was 0.09 in., and for one span loaded 0.15 in. 

Turner-Carter Building Test.—The panels in this building are 17 
ft. 4 in. by 19 ft.6in. The girders are placed in the short direction 
and their dimensions are 10 by 24 in., with two 1-in. square and three 


116 TESTS OF REINFORCED CONCRETE 


Zin, square bars at the middle, and two 1-in. square bars over the 
support. Beams, 7 by 18 in., reinforced with one 1-in. square bar and 
two 2-in. square bars at the middle, and one 1-in. square bar (plus 
ten 3-in. round bars in the slab) over the support, are placed between 
the columns and at one-third points of the girder. The thickness 
of slab is 4 in. 

The structure was designed for a live load of 150 lb. per sq. fis 
and the beams and girders were figured as simply supported, but 
reinforcement was supplied for continuity. The test load was 300 lb. 
per sq. ft. or double the designed load. 

Results of Test—The beams and girders acted as continuous. 
The stresses in steel in the beams were comparatively low, the max- 
‘mum observed for the test load being 11 000 lb. The stresses in 
concrete, however, at the end of the beam reached 1 100 lb. per sq. in. 
At the middle the compression in concrete reached only 350 lb. per 
sq. in., which shows that the. compression there must have distrib- 
uted itself over a large portion of the slab. In the girders the tensile 
stresses at the middle reached only 8 000 lb. per sq. in. At the 
supports no measurements were taken because the steel was not 
accessible. ‘The compressive stress at the end of the beam in the 
bottom was 900 Ib. per sq. in. and was very low at the center in the 
top surface. 

In both beams and girders, the total compression was much 
larger than the total tension, a condition that was found in the 
previous test. As far as observation shows, the entire slab acted as 
compression flange of the T-beam. 

General Conclusions.—In drawing conclusions from tests on com- 
pleted structures, it must be remembered that low stresses in the steel 
do not indicate a large factor of safety. The conditions are the same 
as were explained in connection with laboratory beam tests (see p. 
27) in which the stresses at half the maximum load were small, 
while the maximum load stressed the steel to the elastic limit. The 
results of such tests must be used with caution. 


TESTS OF OCTAGONAL CANTILEVER FLAT SLABS 


An interesting test of cantilever flat slabs supported on a central 
column, as shown in Fig. 40, p. 118, was made in the year 1914 by 
Mr. Edward Smulski under the supervision of Sanford E. Thompson. 

Eight specimen slabs were made octagonal in shape, 6 ft. 6 in. in 


TESTS OF OCTAGONAL CANTILEVER FLAT SLABS Die 


small diameter, and with an octagonal column head in the center 
built monolithic with the slab and having an inside diameter of 2 ft. 
The slab was 4 in. thick. The reinforcement of Specimens 1 to 4, 
arranged as shown in Fig. 40, differed in the diameter of bars used 
for the five outside rings, as shown in the table on p. 120. Speci- 
mens 5 and 6 were similar to 3, except that 10 and 5 radials respec- 
tively were used instead of 20. Specimen 7 was reinforced by four 
layers of bars running in four directions, each layer consisting of 
nine ;%-in. round bars. Specimen 8 was reinforced with steel in 
top and bottom; the tensile reinforcement consisted to two layers 
placed at right angles, with twelve 3-in. round bars per layer, and 
the compressive reinforcement consisted of two layers with eight 
$-In. round bars per layer.*4 

Purpose of Test.—The behavior of a circumferential cantilever 
is similar to the behavior of a flat slab floor system at the support. 
The purpose of the test was to compare the effectiveness of circum- 
ferential reinforcement with that of band reinforcement and to 
determine the most effective distribution of steel between rings and 
radials. The results of the test are directly applicable to flat slab 
construction. 

Materials of Construction—Concrete in the proportion of 
1:2:4 was used. The compressive strength of 6-in. cubes, 
tested at fifty-two days, was 1 100 lb. per sq. in. Reduced to eight 
16-in. cylinders and to twenty-eight days, the strength of the con- 
crete was about 1 400 Ib. per sq. in., or lower than that of first-class 
1:2: 4 conerete. ‘ 

Plain round bars with an average elastic limit of 35 000 lb. per 
sq. In. were used. , 

Method of Testing.—In testing, the slabs were placed on a 
wooden column resting upon a base which distributed the load to 
the soil. The load, consistiag of pig iron averaging 56 lb. per pig, 
was placed on swings arranged along the circumference of the can- 
tilevers as shown in Fig. 40, p. 118. By this method the point of 
application of the load, and therefore the moment arm, was posi- 
tively fixed. | 

Deformation Readings.— Deformations in steel, due to the loading, 


4 Design of specimens 1 to 6 were Smulski system, specimen 7, four-way 
System, and specimen 8, two-way system. Specimens 5 and 6 were of pre- 
liminary type. Their results are not given because they are not comparable 
with other specimens, 


118 


TESTS OF REINFORCED CONCRETE 





Fra. 40.—Cantilever Slab and Loading Platform. (See p. 116.) 


TESTS OF OCTAGONAL CANTILEVER FLAT SLABS 119 


were measured by a Berry Extensometer on 8-in. gage lines. For 


this purpose, gage holes 4 

about z5-in. diameter 2a Bee 
nemaeeee 
lag eal ie i 


were drilled in the steel. 

Each ring and each bar / 

was provided with at ee |p [Re 
least four gage lines to a 
eliminate the possibility 
of erratic results. Aver- 

- age stresses were plotted 
on a deformation dia- | 
gram. Fig. 41, p. 119, 
shows the deformations 












Si NS SS 





= 
& 
@© 





Load in Hundred Pigs (Average Weight of Pig, 56 Lb.) 





at different loadings for 4 
Specimens 1 and 2, and nae 

mig 42, p. 120,: for ; ae ee 
Specimen 7. The curves al Ae a vl Ase, 
for Specimen 8 are sub- 0 10 20 30 40 10 20 0 10 20 
stantially like those for Unit Stress in ridlanal Pounds 


SPECIMEN 2 
Specimen 7. 


Results of the Tests. 
—The results of the tests 
are shown in the table 
on p. 120, which gives 
the dimension of speci- 
mens, total loads, and 
load per pound of ten- 
sile steel. The measured 
stress and the load at. 
first visible crack also 
are given in the table, 
from which it is evident 
that the first visible 
cracks occurred at about 
two-thirds of the load 
at the elastic limit. 
Judging from the stress 0 10 20 30 40 0 10 20 30 40 0 10 20 


Unit Stress in Thousand Pounds 
diagrams, hair cracks SPECIMEN? 


invisible to the eye must Fia. 41.—Deformation Diagrams for Slab Speci- 
have appeared at a mens 1 and 2. (See p. 119.) 


Load in Hundred Pigs (Average Weight of Pig, 56 Lb: ) 





120 TESTS OF REINFORCED CONCRETE 


smaller load corresponding to the break in the deformation 
curve. 


Summary of Results of Tests of Octagonal Cantilever Flat Slabs 


Octagonal slab 6 ft. 6 in. inside diameter; column head 2 ft. diameter; 1:2: 4 


concrete; mild steel. Specimens 1 to 4 Radial; Specimen 7, 4-way; Specimen 
8, 2-way. All slabs 4 inches thick. 


ee 


on moh | Specimen Number 




















=I Number and Diameter ais fo 4 4 a 33 

cr of Round Bars ets aa » : Roo ne 

© : Hg =! We) on sa £ ra] o 

oF in Inches hh o fe) a 2 2 - 

© mee a © 7H te) a7) ae 

rs 65 ee fs a S|: ox Do 

a aey| 35 42 glee melee 235 

5 -6= A (oP) ~ am Aa. +o if 
NM — ne) = o sao s lover tsiitel G 
> eZ A : 5 s ra 2 * a ee Rae Te es 
@ © |Radial|Outside|Straight| {3 . te © = oO Py a) 
Q = j Sno O.8 ep I Qu ars Bo oO 

oD bars | rings bars’ ig.= & . Oh si oe aoe 3 $ 38 Be 
g | 3 s84| s¢ |88| $4 | 22] 85 | gee 
a < < < H 4 4 = 
AT \ 4226 120 —$) 0 "ye lee 29.2 | 0.368t |2.56) 18 800 | 442 12 400 | 14 000 
44. \ 5022) 120 — "215 9 eee 2.2 | 0.484t |2.62| 22500 | 448 | 15 000 | 15 000 
AD FT 2 Ot Sa ae Ulcer 2.2 | 0.901¢ 12.60} 32 000*| 447 | 21 200 | 16 000 
25 (101.2 120 — 4 Sr eee 22 NPI 3s 2.50| 42 500t| 421 | 26 600 | 15 000 
Bt W600 accel aaeeaetete 36 — | 5.5 |........ 2.25| 12600 | 210 8 950 | 15 000 
50 | 56.0 Bee a: 24— 3 Mase pine we QESS 12 BOUT eee 8 900 | 15 000 

Compression |16 — 3 





So | ee ee 
* Estimated from stress diagram. Broken by accident at 29 500 lb., before elastic limit was 
reached. 


+ Estimated from stress diagram. Elastic limit not reached at maximum applied load. 


+ Only part of the area of Ring 5 was considered as effective because it was placed too near 


the column head and therefore carried smaller stress than the other rings. 


Load in Hundred Pigs 
S 


to 












El peels ls] [eal 
7 ea ce 
‘creda Ae eA 


1020.30 40.0 10 20 30 40 50 0 10-20 30 40 0 10 20 30 40 
| | DiDe| Dg Ds 


a] sae | | Beebe [Or] oe Las 
gen FA —— 
S4aenan aa 


if | 
y| roe | Vi) ela ee 
0 10° 390° 30. 40 1507-07 10 20 Fal, e 10 20 30 40 50 0 10 20 30 40 
Unit Stress in Thousand Pounds 


%> 





~— 













SS & 


Fic. 42.—Deformation Diagrams for Slab Specimen 7. (See p. 119.) 


The first crack, at first hardly noticeable, extended all the way 


around the circumference of the column head, several inches from its 
edge. Upon additional loading, this crack opened slowly and 
other circumferential and radial cracks appeared. The test was 


TESTS OF OCTAGONAL CANTILEVER FLAT SLABS 121 


discontinued after the steel had reached the elastic limit with the 
exception of Specimen 4, in which the elastic limit of the slab was not 
reached on account of the difficulty of applying further loading. No 
cracks developed within the column head although the radials were 
stressed to elastic limit and the hooked portions did not bear against 
the center ring. Of interest is the fact that the cracks in Specimens 
7 and 8, reinforced with bands of bars, were also radial and circum- 
ferential. 

From the stress diagrams, it is noticeable that the outside Rings, 
1 to 4 in Specimens 1 to 3, and Rings 1 to 3 in Specimen 4 were 
equally effective in resisting the bending moment, the stresses at 
different loads being almost equal. The stress in Ring 5 was smaller 
than in the other rings, which can be accounted for by the fact that 
the ring was placed too near the column head. ‘The stresses in 
Rings 6 and 7 within the column head were very small, showing that 
very little stress is transferred by the radials to the center rings. 
Evidently, most of it is transferred to concrete by bearing. 

Splicing of Rings.—All rings were spliced with a 50-diameter lap 
During testing, special attention was paid to the behavior of the 
steel at the splices and it was found that the elastic limit was reached 
without any movement being observed at the splices. 

Conclusions.°°—(1) First crack occurred at substantially the same 
measured stresses in the steel, irrespective of the arrangement and 
amount of reinforcement. The load at first crack increased with the 
increase of reinforcement. 

(2) The actual load sustained in all specimens was larger than 
would be expected from ordinary methods of computation, proving 
the effect of Poisson’s ratio. The reduction of bending moment 
coefficients suggested for flat slabs on p. 331 is justified. 

(3) The relative effectiveness of the various arrangements of steel 
can be obtained by comparing the load per pound of tensile steel, 
which for Specimens 1 to 4 varied between 420 Ib. and 450 Ib. and for 
Specimens 7 to 8 between 210 and 225 lb. 

(4) In specimens reinforced by rings, the stresses were uniformly 
distributed over all rings. (See stress diagrams, p. 120.) 

(5) The lap of 50 diameters of a plain bar was sufficient to develop 

the elastic limit of the rings. . 


55 Taken from report by Sanford E. Thompson. 


CHAPTER IV 
THEORY OF REINFORCED CONCRETE 


Reinforced concrete is concrete in which steel or other reinforcing 
metal is imbedded to increase its strength. The reinforcement in 
general consists of bars of small diameter connected with the con- 
crete by bond. Since the reinforcement, has little stiffness in itself, 
in compression members 1t relies upon the concrete and the ties for 
lateral support. When properly designed the reinforcing bars are 
capable of developing full tensile or compressive resistance. Some- 
times rigid structural shapes are used as reinforcement. 

- Not all construction consisting of steel and concrete may be 
classed as reinforced concrete. The function of the two materials 
in a completed structure determines whether it is a reinforced con- 
crete structure or a steel structure fireproofed with concrete. 

An arch of the Melan type, for example, consisting of rigid metal 
ribs capable of carrying some of the load unassisted, but depending 
largely for its stiffness and ultimate strength upon the concrete, 
(see Vol. III) may be considered a reinforced concrete structure. 
A similar arch, consisting of metal ribs connected laterally by metal 
bracing and strong enough to carry the entire load, although encased 
in concrete, cannot be considered as a reinforced concrete structure, 
because the concrete does not serve the purpose of resisting stresses, 
but only the auxiliary purpose of protecting the steel against cor- 
rosion and giving the arch the appearance of a masonry structure. 
Any increase in strength of the structure due to the concrete is 
only incidental. Similarly, conerete columns and other mem- 
bers with structural steel shapes may be considered as reinforced 
concrete when the concrete serves to increase the strength of the 
member. When, however, as in steel frame structures fireproofed 
with concrete, the steel is self-supporting and designed to resist 
the whole of the stresses while the concrete serves chiefly for pro- 
tection, the member is not reinforced concrete. 

The theory of the design of reinforced concrete is definitely estab- 

122 


GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS 123 


lished. The action of the combination of steel and concrete in ten- 
sion, compression, and shear is well understood, so that a thor- 
oughly rational treatment is possible. In practice, in beam design, 
the straight-line theory, as it is termed (see p. 127), which was 
selected and adopted by the authors for the First Edition of this 
treatise, In 1905, has since that time been accepted universally as 
affording the simplest method of computation and as giving results 
which may be used in design with safety and economy. 

In this chapter is presented the analysis of this straight-line 
theory of stresses for rectangular beams (p. 129), followed by the 
application of the same theory to T-beams (p. 133). The analysis of 
a beam with steel in top and bottom is given on p. 137. Wedge- 
shaped beams are treated on p. 140 and unsymmetrical T-beams on p. 
149. The analysis of shear and diagonal tensionwill be found on p. 143. 

The theory of columns of concrete reinforced with vertical steel 
bars is treated on p. 159, and that of columns reinforced with ver- 
tical steel bars and spirals on p. 160. Analyses and formulas are 
presented for the distribution of stresses in reinforced concrete under 
combined thrust and bending moment (p. 164) for use in arch design 
and in the design of columns and beams with eccentric load or 
thrust. 

The theory of members subjected to direct tension or pull, p. 162, 
and also to pull and bending moment is given on p. 189. The theory 
of reinforced concrete chimney design is treated in Chapter XX. 

Formulas for use in practical design, with illustrations of methods 
of treatment, will be found in Chapter V. Tests of reinforced 
concrete, covering all usual features of design, are taken up in Chap- 
ter III. - 


GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS 


Concrete is very strong in compression, but is brittle and unreli- 
able in pull or tension. Therefore, it cannot be used safely nor 
economically where tensile stresses have to be resisted. Steel, on 
the other hand, being a comparatively ductile material, is well 
adapted for resisting pull, but is more costly than concrete for resist- 
ing compression. The economy in the use of reinforced concrete is 
obtained by placing concrete where compressive stresses are to be 
resisted, and steel where tensile stresses are to be resisted. The bond 
and shearing resistance of concrete holds the steel and concrete 
together so that they act as a unit. 


124 THEORY OF REINFORCED CONCRETE 


Requirement for Formulas for Reinforced Concrete Beams.— 
The behavior of reinforced concrete beams under load, as discussed 
on p. 20, is different from that of homogeneous beams. The 
location of the neutral axis for varying intensities of loading is not 
constant. ‘The sum of compression stresses in the concrete is nearly 
proportional to the load, but the tensile stresses in the steel are not 
proportional to the load because of the variable proportion of ten- 
sion resisted by concrete (see p. 97). It is impossible to make 
formulas which would represent actual stress conditions for all loads 
within elastic limit. This, however, is not the purpose of formulas 
used for design. Their purpose is to supply a construction to carry 
a load equal to the design load multiplied by the factor of safety, 
without either steel or concrete exceeding the elastic limit. T his 
being the case, the design formulas must be based on conditions 
existing at the elastic limit. In practice, instead of multiplying the 
design load by the factor of safety and working with stresses at elastic 
limit, the work is simplified by establishing working stresses equal 
to the stresses at elastic limit divided by the factor of safety, and 
using them with bending moments for design load only. It is 
important to keep the purpose and the origin of the formulas con- 
stantly in mind, to avoid using them for purposes for which they 
are not intended. 

Formulas for reinforced concrete beams must satisfy the following 
requirements: | 


(1) The compressive stresses in concrete for working loads 
must not exceed the allowable unit stress; 

(2) The beam must have the required factor of safety based 
on the elastic limit of steel. , 


The first requirement fixes the unit stresses for concrete. 

To satisfy the second requirement, it is necessary in the analysis 
to eliminate the variable amount of tensile stress carried by the con- 
crete (see p. 30), by assuming that all the tensile stresses are resisted 
by the steel. Analysis based on this assumption will not repre- 
sent the actual conditions in a beam under working loads because 
the actual stresses in steel will be less than the computation would 
show, but it will give the required factor of safety and, therefore, 
be correct for design. On p: 27 is given a comparison between 
actual stresses and stresses computed by the accepted formulas 
for beams with different ratios of steel to concrete. It is seen that, 


GENERAL PRINCIPLES OF REINFORCED CONCRETE BEAMS 125 


for earlier stages of loading, the actual stress is much less than the 
computed. ‘This difference, which is due to the tensile resistance of 
concrete, decreases with the increase of the load. At the elastic 
limit of steel, the stage upon which the factor of safety would be 
based, the computed stresses agree fairly well with the actual stresses. 
- The action is shown by the tests, illustrated in Fig. 9, p. 29, 
which give the deformations of concrete and steel at various loads. 
Stresses, of course, are proportional to the deformations. 

Factor of Safety.—By the factor of safety is understood the 
ratio between the load at which the construction would fail and the 
load which it is intended to carry. The words “ fail’ and “ failure ” 
do not mean actual collapse, but a state where the structure ceases 
to be useful. If the material of which the construction is composed is 
equally strong in tension and in compression, the factor of safety is 
the same against both tensile and compression failure. Ina structure 
of a composite material, such as reinforced concrete, the factor of 
safety against compression failure may be different from the factor 
of safety against tensile failure. The factor of safety of the structure 
as a whole is governed by the smaller factor. 

In reinforced concrete construction, steel is considered as the more 
reliable of the two materials, because it is manufactured under 
standardized and uniform conditions and the variation in its elastic 
limit and ultimate strength is, therefore, small. The manufacture 
of concrete, on the other hand, as well as the strength and character- 
istics of the materials composing it, varies from job to job so that the 
ultimate strength of the concrete in different constructions may vary 
to a great extent. Also, during construction, concrete may be 
called upon to resist considerable stresses before its full strength 
has been attained. The allowable unit stresses in compression are 
so selected that the construction has a larger factor of safety against 
compression failure than against tensile failure. The structure is, 
therefore, governed by the factor of safety of the tensile steel. 

The factor of safety, governed by the reinforcement, is based 
upon the elastic limit of the steel and not upon its tensile strength, 
because failure of the beam follows closely after stresses in steel reach 
the elastic limit., The ultimate strength of steel in tension is of no 
significance as far as the strength of reinforced concrete is con- 
cerned. 

Formulas Considering Tensile Strength of Concrete.—In analyz- 
ing the results of the tests in-the early stages of the loading, it is 


126 THEORY OF REINFORCED CONCRETE 


sometimes necessary to consider the tensile stresses in concrete. 
Formulas for such a case are given on p. 140. 


ASSUMPTIONS 


In the analysis of beams, the following assumptions will be made: 


(1) A section that is plane before bending remains plane after 
bending. (See p. 127.) 
(2) Modulus of elasticity of concrete is constant. (See p. 201.) 
(3) Tension is borne entirely by the steeel. (See p. 124.) 
(4) Adhesion of concrete to steel is perfect within the elastic 
limit of the steel. 
(5) There are no initial stresses in the steel. 


Reasons for selecting these assumptions are as follows: 


(a) Beams designed by formulas based on them have the 
required factor of safety. 

(b) The method of design based on them is the simplest. 

(c) They have been adopted by the highest authorities in 
America and Europe. 


Assumption (2) is not actually correct. The modulus of elas- 
ticity varies with the variation in intensity of the stresses; but the 
allowable compression stresses and the modulus of elasticity have 
been selected in such a fashion that the results obtained by the simple 
formulas give the same dimensions as would be obtained with the 
more complicated formulas using the variable modulus of elasticity. 
The maximum fiber stress obtained by formulas is larger than the 
actual stresses. In the selection of the working stresses, this has 
been taken into account. 

Working stresses recommended in this book should not be used 
with other than “ straight-line ”’ formulas. 


ANALYSIS OF RECTANGULAR BEAMS 


Bending Moment and Moment of Resistance,—In a beam sub- 
jected to bending, the bending moment due to the external forces, 
or loads, is resisted by the moment of the internal resisting forces, 
which will be called stresses. Since, by the simple mechanical laws 
of equilibrium, the bending moment must be equal to the resisting 


ANALYSIS OF RECTANGULAR BEAMS 127 


moment of the internal forces or stresses, the unknown stresses in 
the materials may be found by equating the known external bending 
moment to the internal resisting moment. 

Straight-Line Formula.—The stresses cause deformation in the 
material and the consequent deflection of the beam. At any ver- 
tical section through the beam, the compressive stresses above the 
neutral axis cause shortening of the fibers, and the tensile stresses 
below the neutral axis cause lengthening of the fibers. According 
to the first assumption above, a section that is plane before bending 
remains plane after bending. This means that when the beam is 
bent, a plane section through the beam simply swings into a new 
position, without warping. 


ee ee eee Before Bending-------- 
me tiges. pI 27 for Oaihe pads 
example, section AB after f | 
bending assumed the posi- A_ Ay 









tion A’B’ by turningaround 


Compression 


point O at the neutral | nics 
axis. Therefore, the de- Ae OF WO 
formation, or change in \ ee 
length, in any fiber is pro- ‘ 
portional to its distance /__| ney 

from the neutral axis and Ebina pea 98 Puna J 

the variation of deforma- E ie ices ok After Bending 2—-->2--22—<2 

tion of different fibers pyc. 43—Plane Section before and after 
may be represented by a Bending. (See p. 127.) 


straight line. 

According to the second assumption, the modulus of elasticity 
is constant for all intensities of stresses. With a constant modulus 
of elasticity (see p. 126), the ratio of the deformation to the stress is 
constant. ‘Therefore, the variation of the resisting stresses from the 
neutral axis upward can be represented by a straight line, in the 
same manner as the variation in deformations. The compressive 
stresses then form a triangle having for its base the stress in the 
extreme fiber, f., and for its height the distance from the extreme 
fiber to the neutral axis, kd. The total compression may be con- 
sidered as concentrated at the center of gravity of the triangle, which 
Is distant $kd from the extreme fiber. (See Fig. 44, p. 129.) 

According to the third assumption, all tensile stresses are resisted 
by the steel. They may be considered as acting in the center of 
gravity of the steel, which is at the center of the bar if there is only 


128 THEORY OF REINFORCED CON CRETE 


one layer of bars, and at the center of gravity of the set of bars, if 
there is more than one layer. 

For equilibrium, the sum of all forces must equal zero; that is, 
the total compression must be equal to the total pull. The total 
tension and the total compression, which are equal and act in oppe- 
site directions, form a couple with a moment arm equal to the dis- 
tance between the center of steel and the center of gravity of the 
triangle of compressive stresses. ‘The moment caused by this couple 
is the resisting moment. For equilibrium this must equal the 
bending moment due to exterior forces. 

The discrepancy between the actual conditions in a beam and the 
assumption of straight-line distribution of stresses is discussed 
on p. 142. | : 


NOTATION 


For this and succeeding analyses, let 
h = total depth of beam; 
t = thickness of T-beam flange, i.e., thickness of slab; 
b = breadth of rectangular beam or breadth of flange of T-beam; 
b’ = breadth of web of T-beam; 
A, = area of cross section of steel = pbd; 


p = ratio of steel in tension to area of beam, bd 


e ° e A, e 
Pm = Maximum ratio of steel in T-beams, ba for which compres- 


sion stress in concrete equals maximum allowable stress. 


In beams with steel in top and bottom: 


pi = ratio of tensile steel to area of beam, bd; 
ratio of compressive steel to area of beam, bd; 
f, = compressive unit stress in outside fiber of concrete; 
f, = tensile unit stress in outside fiber of concrete; 
f; = tensile unit stress, or pull, in steel; 
; = compressive unit stress in steel; 
E, = modulus of elasticity of concrete; 
E, = modulus of elasticity of steel; 
n= _ = ratio of moduli of elasticity; 
d = depth from outside compressive fiber to center of gravity of | 
steel; . 


Ss 
I 


ANALYSIS OF RECTANGULAR BEAMS 129 


a = ratio of depth of compressive steel to depth, d, of beam; 

k = ratio of depth of neutral axis to effective depth of raty d; 

Z = total amount of shearing stresses in distance, s, and cantik A in 
any horizontal plane below neutral axis. 


| 






Total Compression 


Moment Arm 


Total Tension 
, bd, 
va i pdf; 
n 


Fig. 44.—Resisting Forces in a Reinforced Concrete Beam. (See p. 129.) 


| 
! 
| 
3 
| 
| 
! 
=o es 


kd =depth of neutral axis below the compressive surface in a beam; 
j = ratio of lever arm of resisting couple to depth d; 
jd = moment arm, i.e., distance between centers 6 tension and 
compression ; 
M = moment of resistance, or bending moment in general; 
C= constant i in ‘Tables, pp. 880 and 881. 


Since it is assumed that a section which is plane before bending 
remains plane after bending, we have the proportion 


stretch in steel ms Clk) 
deformation in outside compressive concrete fibers kd 





stress per square inch ff. 


= and unit 


and since unit stretch in steel = chad ain of eld Stik ipmae ns 





stress per square inch __f, 











deformation in concrete = modulus of elasticity = j we have 
fe 
deny be on 
Z = i or ahve Hise Je 
E, 
fe ar 1 
#2 a) era Za ae Ae 


130 THEORY OF REINFORCED CONCRETE 


Solving Formula (1) for fe 


k 
Lee locTaeRy: ale te AS 


Now, as stated above, for equilibrium the total tension in the steel 
must be equal and opposite to the total compression in the concrete. 
The total tension in the steel is its unit stress, fs, multiplied by the 
area of the steel, pbd, and the total compression in the concrete is rep- 
resented by the area of the pressure triangle, 3f.kd, times the breadth 
of the beam, 6. Equating these two forces and canceling out the bd 
which occurs in both, 


of 1, 


If the value of k in Formula (2) be substituted for the k in Formula 
(4) we have 


“ 1 
ods i) 
aE (t+ ag 
For any given percentage of steel the values of f, and f. cannot be 
assumed independently, as they bear a constant ratio to each other. 
Substituting the value of f. in Formula (3) for fc in Formula (4) 
we have 7 


(5) 


k k 
ie ase GER a; _ previo, Gein 


Solving this quadratic equation and adopting the positive sign 
before the square root, 





kz = np + Vinp + (ip) (7) 


From Formula (7) the location of the neutral axis may be deter- 
mined for any percentage of steel, p, and any assumed ratio of mod- 
uli of elasticity, n. Values for k are given in Tables, pp. 880 and 881. 

The center of gravity of the compressive stresses is distant 3kd 
from the top of the beam, so that jd = d — 3kd = (1—4k)d. 

Since the total compression equals the total tension, the moment 
of resistance of the beam may be obtained by multiplying either the 


ANALYSIS OF RECTANGULAR BEAMS 151 


total tension, pbdf; = Asf;, or the total compression, 5febkd, by the 
moment arm jd. 





M 
M eae: A.f.jd ° e (8) and he — A,jd’ ° e (9) 
= fkjbe _ 2h 
Mie pain « (10) and fo= op + + (1) 


For a given quality of concrete and steel, the values of Va! ote 
and j, are constant, so that we may consider the terms Pisj = sfkq 


equal to a constant = - This changes Formulas (8) and (10) to 


By 
M =. R ©or(19) al pCa (13) 


The constant values in Formulas (8) and (10) may also be called 
$049 = pi.) = R. The formula for depth of beam may be also 
expressed by 


d = (14) 


bR’ 
The area of steel may be obtained by solving Formula (8) for As. 

It is 

M 


Sacer SHO Rat Le couse | ELE) 


For balanced design the area of steel equals 
oka DOU, Lik hi Ae el oe tT hy 


where p is the ratio of steel corresponding to the allowable stresses in 
steel and concrete and d is the depth computed on the basis of the 
Same stresses. 

To obtain the total depth of beam, h, a value e must be added to 
the theoretical depth, d. Then, h =d-+e. 

For the use of these formulas in design, see pp. 203 to 214. 


132 THEORY OF REINFORCED CONCRETE 


Values of constants j, k, and p are given on pp. 880 and 881. 

Values of constants C and R are given on pp. 880 and 881. 

Balanced Design of Rectangular Beam.—As is evident from 
Formulas (9) and (11), the stresses of fc and fs are practically inde- 
pendent of each other. The stress in concrete depends upon the 
depth and breadth of beam, while the stress in steel depends upon 
depth of beam and the amount of reinforcement. 

A balanced design of a rectangular beam is one in which the 
stress in steel and the stress in concrete reach their maximum allow- 
able values at the same time. The moment of resistance of concrete, 
figured by Formula (10), is equal to the moment of resistance of steel, 
figured by Formula (8). In a balanced design for fixed stresses 
fs, fe, and n, there is a fixed relation between the area of steel and 
area of concrete. Thus, for a stress in steel, f= 16 000, and stress 
in concrete, fe = 650, the ratio of steel for a balanced design is 
p = 0.0077. 

Unbalanced Design of Rectangular Beam.—An unbalanced 
design of a rectangular beam is one in which either the amount of 
steel or the amount of concrete. is larger than required by the stresses. 
In such a case, the moments of resistance figured for the two mate- 
rials, based on their respective working stresses, will not be equal. 
The available strength of the beam will then be governed by the 
material giving the smaller moment of resistance. 

Thus, if the amount of steel is larger than that obtained from 
the formula As = pbd, where p is the ratio of steel for a balanced 
design corresponding to the selected unit stresses, the design has 
more steel than is required by tensile stresses. Therefore, the bend- 
ing moment must be based upon the compressive strength of the con- 
crete, and the steel cannot be stressed to its full working value. 
For a bending moment based on the full working value of steel, 
the concrete stresses would be excessive. If such a condition should 
occur in a design, the amount of steel should be reduced, unless the 
increase in the amount of steel was necessitated by bond stresses. (See 
p. 264.) 

On the other hand, if for any reason an amount of steel smaller » 
than A, = pbd is used (p being the ratio for a balanced design), 
the moment of resistance must be based on the reinforcement. 
The stresses in concrete will be smaller than the allowable fiber 
stresses. To stress the concrete to its working value would over- 
stress the steel. | 


FORMULAS FOR T-BEAMS 133 


FORMULAS FOR T-BEAMS 


If a reinforced concrete beam is built monolithic with the slab, 
the beam may be considered as a T-beam in which a portion of the 
slab acts as a flange. (See Fig. 45, p. 133.) 

The formulas for T-beams given below are based on the same 
assumptions as those for rectangular beams. Tension is considered 
as taken entirely by the steel, and the variation of stresses in con- 
crete is according to a straight-line. Unless the slab is very thick, 
the neutral axis is located below the flange. 

For notation, see p. 128. 


Case I.—Neutral Axis below Flange, kd > t. 
Formulas Neglecting Compression below Flange.—In ordinary 
T-beams, the amount of compression stresses resisted by stem below 


the flange is insignificant and may be neglected. This simplifies 
the formulas without causing any appreciable error. 







Neutral 
Axis 


ae 





Fig. 45.—Resisting Forces in T-shaped Section of Beam. (See p. 133.) 


The location of the neutral axis is determined in the same man- 
ner as for rectangular beams. The ratio k may be expressed by 


ae 1 


1+ 4 





(17) 


The total tension is equal to the unit stress in steel, f,, multiplied 
by the area of steel, A,. The total compression in the concrete is 


; : kd -— 
represented by a trapezoid, the sides of which are f, and fou s 


a 


134 ‘THEORY OF REINFORCED CONCRETE 


and the depth is equal to t. The total compression, therefore, equals 


2k 5 
2kd — t 
Vic kd bt, also ‘eo imi bt. 





By equating total tension to total compression acting on the 


section 
Dh st 
Asfs came” v Id bt. ° ° e ° ° ° ° (18) 





Solving the two above equations for kd and eliminating f. and fs, 
we get 
Position of neutral axis 














4 2nd A, + bt? 
Keo As oe (19) 
The distance of the center of compression from upper surface of 
beam 
3kd — 2t/t 
,- MAG). ictroa a’ se aU otea Re (20) 
Arm of resisting couple 
90. chi 
Moment of resistance 
Qkd — 1, ,. 
Meat, 70fe! i) Bree (21) and M= Phd btjdf.. (21a) 
Fiber stresses 
M 
L-a oe 
Mkd » hci 
fe = bikd — Bhd (2); sg beter ceg, PE 
Area of steel 
M 
A= 2 
fajd ce 


The form of this equation is the same as for rectangular beams 
(Formula (15) p. 131). The difference is in the value of 7. As in 
the case of rectangular beams, the moment of resistance of a T-beam 
may be governed either by the allowable tensile stresses in the rein- 
forcement or by the allowable compressive stresses in concrete, 
whichever gives the smaller value. For a T-beam of fixed dimen- 


FORMULAS FOR T-BEAMS 135 


sions there is one value of p for which the stresses in steel and con- 
crete reach their allowable stresses simultaneously. This is the 
maximum ratio of steel that can be fully utilized without over- 
stressing the concrete. Its value is given in the diagram, p. 894, 
for different unit stresses and different dimensions of the beam. 
Minimum Depth of T-Beam.—In a T-beam design, the depth 
of slab to be used for the flange is designed first. This fixes the 
value ¢. ‘The width of the flange depends largely upon the thick- 
ness (see p. 217), so that for fixed ¢, the value 6 is practically fixed. 
Since the area of compression flange is fixed, for specified working 
stress in concrete, fc, the available compression is practically fixed. 
The moment of resistance being equal to total compression multiplied 
by moment arm jd, it is clear that with total compression practically 
fixed there corresponds to each moment M, a definite value d, for 
which the stresses in the flange are equal to the maximum working 
stresses. This depth is called minimum depth. The formula for 
minimum depth is 
Minimum d = a ae 
2k — 5p 
q)ae 


Calling Ca = ( oy 


, the formula becomes 


Minimum depth 
M 


Minimum d = Cot 


Ho (20) 


Values of Cz depend upon the ratio by and upon, k, 7 and f., which are 
d 


functions of the allowable working stresses. The diagram on p. 894 
gives values of Ca for various assumptions. 

Maximum Area of Steel in T-Beam Permitted by Compression.— 
In a T-beam of definite design, the compression area is fixed as 
explained above. For a definite depth d and working stresses f., f, 


siotbak 


d 
ay 
The total tension available equals A,f., in which only f, is fixed. 
The available moment of resistance of the beam equals either total 
tension or total compression multiplied by the moment arm, which- 


and n, this fixes the available total compression, which is f, 


136 THEORY OF REINFORCED CONCRETE 


ever gives the smaller value. If the total available tension, A.fs, is 
smaller than the total available compression in the flange, then the 
tension governs the bending moment and it is M=A,f,jd. When the 
total available tension, A.f., is larger than the available total com- 
pression in the flange, the compression governs the bending and the 
area of steel is not fully utilized. It follows that, in a rational 
design, such an area of steel should be used that the total tension is 
smaller than, or equal to, the total compression. 

The maximum area of steel which can be used economically is 
that for which the total tension equals the total compression, 


of 


Maximum A,f, = fo oot 


Calling Maximum A, = Pmbd in substituting in the above equation, 
we have 


2k — 5 
pnbdf, = f, es es 
or finally 
Maximum ratio of steel | 
t 
2k— 5 
Pn = ope oot . > See 
and 
Maximum area of steel 
Maximum A, = Dubd: 2. ae (27) 


The ratio pm depends upon the values k, f, and f., all of which 
: ne : 
depend upon the working stresses and the ratio a The diagram on 


p. 894 gives values of pm for various assumptions. 

Maximum Moment of Resistance of T-beam.—In a T-beam of 
definite concrete dimensions, the maximum moment of resistance is 
the one for which the compression stresses in the extreme fiber in 
concrete reach the maximum working value. 





2k — 5 
Maximum M = ah af bid. 
(2% — 5)it 
Calling, as on p. 135, Ca = Remy. the formula becomes 


Maximum Mo = dCqbt- 2) (28) 


BEAMS WITH STEEL IN TOP AND BOTTOM 137 


The diagram on p. 894 gives values of Cg for various assum ptions. 
Formulas Considering Compression below Flange.—For large 
beams where the stem forms a large part of the compression area, 
the above formulas do not give results accurate enough for practical 
purposes. or such cases, formulas given below may be used, which 
take into account the compressive stresses in the stem as well as 
in the flange. The following formulas are derived by the same prin- 
ciples used in derivation of formulas in the previous analysis. 
Depth to neutral axis 
2ndA, + (b — b’ en) nA,+(b—0b’)t\? nA,+(b—d't 
b’ +( b’ ore b/ (29) 
_ (kdi? — 3t°)b + [(kd — £)2(t + 4(kd — £))]b" 
‘eo t(2kd — t)b + (kd — 2)?’b 


kd = 








(30) 


Arm of resisting couple 
De ee Rr es Boge Nc ee et (3.1) 
Moment of resistance 
Pe Aad. (32) Me sgl ht — bt -+ (kd — £)2B’]jd. (33) 
Fiber stresses 


M cAal 2Mkd 
eee ee Odie = op ‘bt + (kd — 8)2b"Vjd" 


(35) 

A simplified. method is given on p. 224. This can be used in 
design for determining dimensions of the beam for known stresses 
in steel and concrete. It cannot be used for computing stresses, how- 
ever. 

Case II.—Neutral Axis in Flange or at Underside of Flange, 
kd < t. 

In this case, which occurs only with slabs very thick in pro- 
portion to the depth of the beam, use the rectangular beam formula, 
considering the T-beam as a rectangular beam of the same depth, 
the breadth of which is the breadth of the flange. The percentage 
is then based on the total area, bd. 


REINFORCED CONCRETE BEAMS WITH STEEL IN TOP AND BOTTOM 


In beams reinforced with steel placed both in the compressive 
and tensile portions of the beam, the steel in the compressive portion, 
as in reinforced concrete columns (p. 159), may be considered as 
resisting compression stresses equal to the stresses in the concrete 


138 THEORY OF REINFORCED CONCRETE 


which it replaces, multiplied by the ratio of moduli of elasticity, 7. 
Same assumptions are made as for simple rectangular beams (p. 126). 

The tension in concrete may be neglected, as in beams without 
compressive steel, and all the tension may be considered as resisted _ 
by the bottom steel. Referring to Fig. 46, p. 188, the total com- 
pression consists of the compressive stresses in concrete represented 
by a triangle and the compressive stress in steel. The compressive 
unit stress in steel equals the unit stress in concrete at the same level, 
multiplied by the ratio of their moduli of elasticity. 







ompression in Compression Reinforcement, 


“f-> -C ment 
7 ‘Steel p'bdfg ; 












M. Arm for Comp 


Compression in | 


Concreté rah kbd | 


--d (1-4}-- 
mp.\Steel 


aS (-a)- JES 


M.Arm for Co 















; p, bd fs 
Py bd -J->| Total Tension 


Tensile y 
Reinforcement” 


Fic. 46.—Resisting Forces with Steel in Top and Bottom of Beam. (See p. 1388.) 


The sum of all the horizontal stresses acting on a cross section 
must equal zero;! therefore, the total tension in steel must be equal 
to the compression in concrete plus the compression in the top steel. 
The resisting moment, that is, the moment of the internal stresses, 
may be obtained either by multiplying the total tension or com- 
pression by the distance between center of tension and center of 
compression, or by taking moments about the center of tension steel, ° 
the center of compression in concrete, or the center of compression 
in steel. The moments of rec'stance obtained by either of the four 
methods must be equal. To find the stresses for a certain loading, 
the moment of resistance taken in any one of these ways is equated 
to the known bending moment. 

Formulas.—Deformations, as usual, are assumed to vary directly 
as distance from neutral axis, hence from Fig. 46, using notation 
on p. 138, 


dda Le ee 
Ey OUR eh ae Whence k = 1 7 
Ste dk k 1+ pis 
E, nf c 
1 This is a law of simple statics. Otherwise there would be a movement in 
the beam, 





(36) 


BEAMS WITH STEEL IN TOP AND BOTTOM 139 


By comparing the above equation for k with that given for simple 
fs 
nfo’ 


neutral axis is the same irrespective of whether the beam is provided 
_ with compressive steel or not. 

By similarity of triangles in Fig. 46, p. 138, the following relations 
between the unit stresses may be obtained: 


beams, p. 129, it is evident that for any ratio of —, the position of the 


f's 





(37) and fi =inf, 








{, = op (39) and ie Hide Sere hy 


The total tension in steel equals bdp,f, and the total compression 
in steel and concrete is 


bd + dap'f’, = df. + vf). 


Since the sum of all the stresses must equal zero, the total compression 
acting on the cross section of the beam equals the total tension, or 


ba( nip J .)= bdpifs. 


ma / fo Poel ffeil tk? ka 
Pi =7(5 rays} = 7(# eeen: oo De makige is ‘) 
kK ka 
aOR —R\at Oe h: 
Solving equation (41) for k, | 
k = V2n(pi + pa) + 21 + p')? — n(pit yp’). . (42) 


Taking moments about the center of compressive stress in the 
steel, we have 


A ba?| fart —a)— BG - a) | 
or by eliminating f. 


| pill — a)2n(1 — k) — ae — : 
M-—-f,bd? ius fr CAs eee .. (48) 


Whence 


Hence 
(41) 


From which 
es M 6n(1 — k) 
4 (bP 6npi(1 — k)(1 — a) — k2(k — 83a) 





(44) 


140 THEORY OF REINFORCED CONCRETE 


By substituting this value of f, in equations for fc and f’, respec- 
tively, we get 


_M oh 
Se = 5@2 @npid = EC — a) — Ph — 3a)’ eo 
and 
ell. 6n(k — a) 
eat 6npidl — k)(1 — a) — KAR — 3a)” ee 


It may be noted that the denominator in the three above equa- 
tions is the same. A simplified method for designing beams with 
steel in top and bottom 1s given in the chapter on Design. Tables 
of constants are given on pp. 904 to 910. 


WEDGE SHAPED RECTANGULAR BEAMS 


Under this heading come beams in which one or both faces are 
not at right angles to the section of bending moments. They are 
found in retaining walls, footings, dams and also at the support of 
beams provided with haunches. 

When the angle of inclination of the faces with the horizontal does 
not exceed 10 degrees, ordinary formulas may be used. For larger 
angles the following formulas suggested by Professor William Cain ** 
will give closer results. These formulas are approximate. The 
number of experiments made on wedge shaped beams is very small 
so that it is not possible to check the accuracy of the formulas. 

Figure 47 represents part of a wedge shaped beam, both faces of 
which are sloping. The stresses produced by a bending moment 
acting on section AB will not be perpendicular to the section. In 
the steel they act in the direction of the reinforcement, while in 
concrete they are parallel to the compression surface. The direc- 
tion of stresses is marked in the figure. To get a moment of resist- 
ance acting on section AB ‘t is necessary to take components of 
compression and tension at right angles to the section. Following 
formulas are based on condition represented in Fig. 47, p. 141. 

Netaticn.—See pp. 128, 246 and 262. 

Both Faces Inclined. 


cos B 


[oes 
COS? a 





lee np + 2m mes + inp)? | vo eae 


14 See Professor William Cain “Barth Pressure, Retaining Walls and Bins.” 


STEEL IN BOTTOM OF BEAMS 141 


hs NE i ecg (48) 
ee M 

° A, cos Bjd * 
Fae 

°  kjbd? cos? a 
_ ple costa 
(tat if, bcos Bi” 


(49) 


(50) 





(51) 


_ V _ M(tana + tan 8) (52) 








bjd bjd? 
mane nA * M (tana + tan 8) 
u= Sojd Sop Aree User ee {(O8) 






Component 
1 
“a bkd cos*x 
bkd cos « 3 


A, fy cos 3 


Component 





Part Elevation Section 


Fig. 47.—Resisting Forces in Wedge Shaped Beam. (See p. 140.) 


Compression Face Inclined. Steel at Right Angles to Section.— 
p= 0°. 





= <3 (— mp + V2np cos? a + (np)?) te tos) 
M 
ee ee ee ey ean) 55) 
2M 


CMa ye ce ey ae tO) 


A cor? oe. EEE are tee are seen Mier dy Lelia 


fe 
V Mtanea 

Cpa ia oe 
V M tana (59) 





aaa Lojd  Sojd? ° 


142 THEORY OF REINFORCED CONCRETE 


T-BEAMS WITH FLANGES ON ONE SIDE ONLY 


T-beams with flange on one side only are found where the slab 
extends only on one side of the beam as at the wall and at openings 
in the floor. It is common practice to treat these beams in the same 
manner as symmetrical T-beams, assuming that the neutral axis 
will be parallel to the compression flange and that the distribution 
of stresses will be the same as accepted for symmetrical T-beams. 
Tests by Professor Bach on Unsymmetrical Beams prove that the 
neutral axis, instead of being parallel to the flange, slopes upward 
toward the side provided with the flange and may even intersect 
the flange. The maximum stresses in concrete occur at the rec- 
tangular corners. The ‘stresses in steel are not affected much by the 
unsymmetrical arrangement of the flange. 

While formulas have been proposed for this condition,” there are 
not sufficient data to warrant their adaption. Until additional 
information is obtained the stresses should be figured in the same 
manner as for symmetrical beams, except that the allowable working 
stresses in compression should be reduced by about 30 per cent. 


COMPARISON OF STRAIGHT-LINE FORMULAS WITH ACTUAL 
CONDITIONS 


Tests prove that the assumption (2) on p. 126, that the modulus 
of elasticity of concrete is constant for all stresses, is not correct. 
Actually the modulus of elasticity is a maximum for small stresses 
and decreases with the increase of the intensity of the stresses. 
(See Volume II.) 

If a plane section before bending remains plane after bending, 
the deformations vary according to a straight line, as shown in 
Fig. 48a. The unit stresses are equal the deformation multiplied 
by the modulus of elasticity. For constant modulus the variation 
of stresses will be the same as the variation of deformations. This 
gives the “ straight-line ” formula, Fig. 480. 

Actually the variation in stresses is different from the straight- 
line variation, because, to get the stresses, the deformations must be 
multiplied not by a constant modulus of elasticity but a varying 
modulus of elasticity, which is a maximum for small stresses near 


10 Gog Karl Hager ‘‘ Vorlesungen wber Theorie des Eisenbetons.”’ 


SHEARING STRESSES IN A BEAM OR SLAB 143 


the neutral axis and decreases for larger stresses. Figure 48¢ 
shows parabolic distribution of stress, as proposed by Prof. Arthur N. 
Talbot. Special attention is called to the fact that all formulas are 
based upon the law that plane section before bending remains 
plane after bending. In the parabolic formula it is not the plane 
that assumes a curved shape, but the variation of stresses acting 
upon it. 

Since in the straight-line formula an average value for the modulus 
of elasticity is used, the actual stresses are larger for the small stresses 
(near the neutral axis) than obtained from straight-line formulas. 
For stresses near the extreme fiber the actual stresses (based on 


Deformations, Stress s Deformation x Constant Ee Stress-= Deformation x Variable E', 
pe i 

Variation in 

Modulus Eg 





Tension Tension 





a ay 
a - bis aplane before bending Straight - Line Parabolic 
a, b,is same plane after bending Variation of Stress Variation of Stress 
(a) (0) (c) 


Fic. 48.—Different Assumptions as to Variation of Stresses. (See p. 142.) 


varying modulus of elasticity) are smaller than in the straight-ling 
formula because the actual modulus of elasticity is smaller than the 
assumed. ‘To compensate for this, the allowable working ‘stresses 
to be used with straight-line formulas were made larger than would 
be permissible for exact formulas. In formulas with parabolic dis- 
tribution of stress the working stresses for straight-line formulas 
would give unsafe results. 


SHEARING STRESSES IN A BEAM OR SLAB 


The bending of a beam produces a tendency of the particles to 
slide upon each other. 

This tendency is called shear. The resistance to sliding offered 
by the material, called shearing strength, is a property of the mate- 
nal. The effect produced by bending is called shearing stress and 
its magnitude depends upon the external loads and upon the dimen- 
sions of the beam. The shearing stresses are produced by external 


144 THEORY OF REINFORCED CONCRETE 


shear. This depends only upon the magnitude and character of the 
loading and is independent of the dimensions of the beam. These 
terms are used in the discussion below. 

In a beam, shearing stresses are developed in two principal direc- 
tions. It is, therefore, necessary to study 

(1) vertical shearing stresses, 

(2) horizontal shearing stresses. 

Vertical and Horizontal Shearing Stresses.—Vertical and hori- 
yontal shearing stresses are equal. They may be computed from 
formulas on page 149. The resistance of concrete to direct shear is 
large. (See Vol. II.) Itis safe to use a working shearing stress of 
at least 200 Ib. per sq. in. Therefore, a concrete girder, beam, or 
slab will always have sufficient area of section to withstand this 
direct shearing stress. However, since the direct shearing stress is a 
measure of the diagonal tension (see p. 148), which is excessive when 
the direct shearing stress is comparatively low, it must always be 
computed in a beam or girder for use in the designing of web rein- 
forcement. 

Magnitude of External Shear.—The external shear is a maximum 
at the support, where it is equal to the reaction. For simply sup- 
ported beams, the external shear may be determined by statics. 
In continuous beams, the actual external shears differ from the 
‘statie shears, as is evident from the formulas given in the section 
on continuous beams. While, with uniform or symmetrical loading, 
the reaction, and therefore the maximum shear, in simply supported 
beams is one-half the total load upon the beam, it will be noticed 
from the diagrams that in the end spans of continuous beams with 
free ends the external shear at the first support away from the end 
may be 25 per cent greater than the static shear, and should be spe- 
cially provided for in cases where the full live load is likely to be 
constantly maintained. For complete treatment of external shears 
‘n continuous beams see Volume Il of this treatise. 

Longitudinal Shear in T-beam.—The projections of the flange of : 
a T-beam on each side of the stem are subject to compressive stresses. 
These are zero at the support or at the point of inflection and increase 
with the increase in bending moment. T he increase in compression 
stresses in the projections of the flange must be transferred from sec- 
tion to section by the resistance to shear at the junction of the flange 
projections and the stem. These shear sections are marked AB and 
A,B; in Fig. 49, p. 145. When the shearing stresses are larger than 


SHEARING STRESSES IN A BEAM OR SLAB 145 


the resistance to shear, the flanges become separated from the 
stem, the compression area becomes reduced, and the beam fails by 
crushing of the stem. Such failures are described in the chapter 
on Tests, p. 36. An analogy may be found between the shearing 
stresses along the flanges of a T-beam and the shear in the rivets of 
a built-up steel beam, which connect the flange angles to the steel 
web plate. 

The magnitude of the shearing stresses may be found in the fol- 
lowing manner: Consider two vertical cross sections of a beam, 
one inch apart. Let M, be the moment at the section nearer the sup- 
port, M2 the moment at the other section, and V the external shear, 
and assume for the sake of simplicity that there is no loading between 


eens a — 









Fig. 49.—Longitudinal Shear in T-beam. (See p. 144.) 


the two sections. Then, from simple laws of statics, the moment 
M2=M,+VX1. The total amount of compression stresses 
produced at each of the two sections is Ms and ike = Ms + Me 

qd ad Jdaks 44 
If the part of the beam between the two vertical sections is detached 
from the beam, the compression stresses act in opposite direction, 
as shown in the figure. Since the amount of compression on the two 
sections is not equal, only a part of the compression stresses will 
balance. The unbalanced difference between the compression 
stresses at the two vertical sections produces shear in the beam. 
This difference is Mz — M, = a This increment of compression 
stresses acts on the whole flange. To get the amount of compression 
on the projections only, it is necessary to multiply the total incre- 
ment by the ratio of the width of the projections b — b’ to the width 
of flange, b. Thus the amount of compression stresses producing 


146 THEORY OF REINFORCED CONCRETE 


longitudinal shear at the juncture of the flange projections and the 
stem equals ree ey) 
Ate Ue 
The intensity of the compression stresses on the flange is not 
uniform, but is a maximum at the extreme fiber and a minimum at 
the bottom of the flange. The unit increment producing shear is 
proportional to the compression unit stresses; therefore, the longi- 
tudinal shear will be a maximum at the top and a minimum at the 
bottom of the flange. If the shearing stress at the top is called v,, 
the minimum shearing stress at the bottom is u a : The total 
shear on two sections AB and Ai Bi, for a depth of flange equal 
t 
2k — = 
to t 1s a(o, + bile = *) 2 = ppt 
kd k 
stresses producing shear, found above, 














By equating this to the 


t 
ab dt Vib ee 


poe Gd tbe 





Val 


from which maximum longitudinal shearing stress in flange 


Dagieiiies” 
(aaa ijd ‘ b t « ° ° ° ° e (60) 
2h — 5 


Vh 





From this formula it is evident that the shearing stresses increase 
with the ratio of the projections to the total width of flange. 

Ordinarily, the shearing stresses are so small that the concrete 
alone can resist them. Excessive longitudinal shear may be devel- 
oped only in short beams with heavy loads. 

As evident from the chapter on Tests, the strength of a T-beam 
is increased by fillets at the junction of the slab and the stem. 
Also, cross bars placed across the beam near the top of the slab 
increase the strength of the T-beam. Where the tensile reinforce- 
ment in a slab is placed across the beam, it strengthens the connec- 
tion between the stem and the flange. Where the slab reinforce- 
ment is parallel to the beam, special short bars should be placed 
across the beam. ‘These serve two purposes: first, they strengthen 
the connection between the stem and the flange; second, they resist 
negative bending moment stresses which are developed by the load 


& 


DIAGONAL TENSION 147 


in the slab. Without the tensile bars, tensile cracks often form at the 
junction of the flange and the stem and materially reduce the effect- 
iveness of the flanges. 


DIAGONAL TENSION 


In any beam, besides direct horizontal tension and compression 
and direct horizontal and vertical shearing stresses, there exist also 
stresses acting in diagonal directions. The maximum diagonal 
stress composed of the tension and the shearing stresses is called 
diagonal tension. 

In steel and other homogeneous beams, diagonal stresses need 
no attention. In reinforced concrete, however, it has been shown in 
beams tested to destruction that, beside tensile cracks at the points 
of maximum moment, diagonal cracks, caused by diagonal tension, 
develop near the supports. (See tests on pp. 38 to 46.) These 
cracks have often been the cause of failure, frequently without warn- 
ing, especially in beams reinforced with straight bars only or pro- 
vided with insufficient web reinforcement. The need of effective 
web reinforcement is discussed in the paragraphs which follow. 

Diagonal Tension in Homogeneous Beams.—The magnitude 
and inclination of the diagonal tension in homogeneous beams may 
be found from the following formula: 


Let fa = diagonal tensile unit stress; 
f: = horizontal tensile unit stress; 
v = horizontal or vertical shearing unit stress. 
Then 2 
Meat iy tere ee aT 


The direction of this diagonal tension makes an angle with the 


ee 
horizontal equal to one-half the angle whose cotangent is sh 


As is evident from the formula, the value of diagonal tension 
depends upon the values of f, and v. Since the values in f, and v 
vary in different parts of the cross section of the beam, the value of 
the diagonal tension and its angle: of inclination also vary. At the 
bottom of the section, where v equals 0, fa equals f: and acts hori- 
zontally. At the neutral axis, the direct tension, fi, equals 0, 


* For derivation, see Merriman’s “Mechanics of Materials,” 1905 edition, 
p. 265. 


148 THEORY OF REINFORCED CONCRETE 


which reduces the formula to fa = v and the angle of inclination 
to 45°. 
Measure of Diagonal Tension for Reinforced Concrete Beams.— 
In homogeneous beams, the diagonal stresses can be determined 
easily by means of Formula 63, p. 149. In reinforced concrete, how- 
ever, the diagonal stresses cannot be computed with exactness because, 
as may be seen from the formula, they depend upon the horizontal 
tensile stresses in concrete, fi. The action of concrete in tension 
cannot be computed. It varies for different stages of loading. Tor 
heavier loadings, concrete cracks, thus decreasing the tensile stresses 
carried by it. The tensile strength of concrete, which may be dis- 
regarded in figuring the moment of resistance of the beam, affects 
the magnitude of the diagonal tension to a great extent, especially 
near the ends of simply supported beams, where the stresses due to 
the bending moment are low and the tensile stresses in concrete 
may not exceed its breaking strength in tension. While the exact 
determination of diagonal tension is impossible, tests show that the 
shearing ‘unit stress, figured as given on p. 149, may be accepted 
as a convenient measure of diagonal tension; that is, the diagonal 
tension may be assumed as proportional to the direct shearing stress. 
Therefore, by adopting proper working stresses based on tests pro- 
ducing diagonal tension failures, formulas for shearing stresses may 
be used for diagonal tension. This measure has been universally 
accepted and, in subsequent discussion, diagonal tension is expressed 
in terms of shearing stresses. | 
Formulas for Shearing Stresses and Diagonal Tension.—A 
convenient and safe method of determining the diagonal tension is 
by accepting for its measure the unit shearing stress as discussed 
above. | 
Let V = total external shear at section considered. (Reaction 
minus the loads between the support and the section.) 
» = horizontal (or vertical) shearing unit stress at section 
considered ; 
b = breadth of beam; 
b’ = breadth of web of T-beam; 
jd = moment arm or distance between center of compression 
and center of tension (approximately, in a T-beam, 
distance between center of slab and steel). 
7, = total shearing stress or diagonal tension in a given length 
of beam, s. 


DIAGONAL TENSION 149 


The following general principles and formulas are discussed in 
the paragraphs which follow: 

(1) Vertical shearing unit stress is equal to the horizontal shear- 
ing unit stress, and acts at right angles to the plane of horizontal 
shearing stress. The distribution of vertical shearing stress over a 
vertical section is shown in Fig. 50, p. 150. 

(2) Horizontal (or vertical) shearing stress is zero at the top of 
the section and changes according to a parabola till it reaches its 
maximum at the neutral axis. (See Fig. 50.) 

(3) If tension in concrete is neglected, the horizontal (or vertical) 
shearing stress is constant below the neutral axis. 

(4) The total amount, then, of horizontal (or vertical) shearing 
stresses developed below the neutral axis at any horizontal plane in 
a distance, s, and width, b is 


Z eae aii « Gite Be peak med) vane ee” (62) 


(5) Shearing unit stress, the measure of diagonal tension, is total 
horizontal shearing stress, Z, divided by the horizontal area, b X s, 


7 
that is v = Vs + bs, 
ad 


Hence 


(63) For T-beams, v = at ire tt BA) 
If the external shear, V, changes in the distance, s, the same for- 
mulas may be used except that V in the formula is the average shear 
in that distance. 
| (6) Diagonal tension may be expressed in terms of the shearing 
stress and the above formulas may be accepted as its measure. 

(7) If the width of the section below the neutral axis is not con- 
stant, the shearing unit stress will vary with the width, 6. The 
smallest b must be taken in figuring maximum shearing unit stress and 
the maximum diagonal tension. 

(8) In continuous T-beams near the support, the maximum 
shearing unit stress will be in the stem directly under the flange. 
The shearing stress and diagonal tension in the plane of tension 
steel (at top of beam) is small, because the width, b, being the total 
width of the flange, is large. 


150 THEORY OF REINFORCED CONCRETE 


The action of horizontal shearing stresses is illustrated in Fig. 50, 
in which is shown a section of a beam between two vertical sections 
spaced a distance s apart. If the bending moment at the left is M,, 
the external shear V, and the distance between the sections under 
consideration s, then, from the principles of mechanics, the bending 
moment at the right is M, = M, + Vs. 

Since the bending moment at the right is larger than the bending 
moment at the left, the unit compressive and tensile stresses at 
the right section are larger than at the left. Consider an arbitrary 
longitudinal plane, ef, above the neutral axis. The compressive 
stresses above this plane are represented by the shaded portions of 
the triangles. They act in opposite directions. At the left they are 


ylt A Cyl hi— hi 
eee 
/ Neutral Axis___ “a 
\ Mt, 
= 
seed 


oe Section Distribution of 
shearing stresses 






Elevation 


Fic. 50.—Horizontal and Vertical Shearing Stresses in Beam. (See p. 150.) 


equal to Cy, and at the right to Cy = Cn+AC,. The difference 
between C,, and Cyr, which act in opposite directions, is AC je. jethis 
tends to move the upper portion of the beam along the plane eff 1€1, 
but is kept in equilibrium by the horizontal shearing resistance in 
the beam on that plane. The shearing unit stress produced by the 
force AC; is equal to the force AC, divided by the area, bs, of the 
plane, effier. 

At the top of the beam, the value of AC,, and also the total shear- 
ing stress is zero; it increases steadily according to a parabola till it 
reaches its maximum at the neutral axis. There its value equals the 
difference between the total compression on the right and the total 
compression on the left. From the ordinary beam formulas, p. 130, 
we know that the total compression may be found by dividing the 


DIAGONAL TENSION 151 


bending moment by the moment arm: thus, at the left, the total. 


: : bald M. ast M, eee Vs 
compression is C, = id and at the right, C, = hiv Maia di 
The difference between C, and C, is 1 Therefore, the total amount 


jd 
of horizontal shearing stress at the neutral axis for the length, s 


+r 


and width, b, is Z = a This has to be resisted by the horizontal 


plane of the beam, bs, so that the shearing unit stress, v, equals Z 
divided by bs. It is 
Shearing Unit Stress 


AG 


Diz ai (65) 


So 


If there is no tension in concrete, the difference between the 
stresses acting above any plane located below the neutral axis is the 
same as the difference at the neutral axis. Consequently, the sum 
of horizontal shearing stresses is uniform at all planes below the 
neutral axis. As the shearing unit stress depends upon the width, b, 
it is constant for rectangular sections, but various for sections with 
variable widths, b. 

At the plane of reinforcement, the stresses in steel at the left are 


Sie M M. _M, V Ss 
t= ae and at the viel ates Shee ns ad auld and the difference, 
T, — T, = = This shows that the total horizontal shear, or the 


tendency to move the upper portion of the beam, is the same at the 
plane of the bars as at the neutral axis. 

If there is tension in concrete, the total horizontal shearing 
stress, Z, on any plane below the neutral axis will be decreased by 
the difference in tension at the two vertical sections above that plane. 


r 


Bond Stresses.—The increase in the stress in steel, equal to ea 


between the two sections considered, must be transferred from the 
beam to the steel. Therefore, bond must exist between steel and 
concrete, or else the beam will slide on the steel instead of increasing 
its stress. Tests of bond, or resistance to slipping of bars, are 
treated on p. 56. Discussion of bond stresses and their importance 
is given on p. 260. 


152 THEORY OF REINFORCED CONCRETE 


Diagonal Tension Acting on an Element of a Beam.—Figure 51 
represents the stresses to which any element-of a beam is subjected. 
In Fig. 51 is shown a rectangular element of the beam, the sides of 
which are dz and dy. This element is kept in equilibrium by six 
forces: two forces, fidy, acting in opposite directions and being either 
direct tension or compression; and four shearing stresses, vdx and 
vdy, respectively, caused by the increment of the bending moment, 
as explained in the preceding paragraphs. The two horizontal 
shearing stresses, vdv, form a couple, vdx X dy, which is resisted 
by a vertical couple, ody X dx. The moments of the two couples are 





Fic. 51.—Stresses Acting on an Element of Beam. (See p. 152.) 


equal; therefore, the horizontal shearing unit stress must be equal 
to the vertical shearing unit stress. 

Illustration of Action of Web Reinforcement.— Figure 52, 
p. 153, illustrates the action of vertical stirrups in a simply sup- 
ported beam. Stirrups do not act until minute cracks open. After 
a, crack forms, as in the figure, the reaction and the shear, V, tend 
to widen the crack and cause failure of the beam. This tendency 
is resisted by the stirrups, acting in tension. Figure 52 also rep- 
resents what would happen if there were no web reinforcement. 
Figure 53 represents the action of stirrups in-a continuous beam. 
It may be noticed that the stirrup gets its stress at the top instead 
of at the bottom. , 

Distribution of Diagonal Tension to Concrete and. Stirrups.— 
Tests prove that in beams with web reinforcement, the diagonal 


DIAGONAL TENSION 153 
tension found by Formula (65), p. 151, is resisted by both concrete 
and steel. The relative proportion of stress taken by the concrete 


_ Ends of Stirrups must be 
late / /“ Anchored in Compression Zone 








— External Shear 





~~ Direction of Force 
Acting on Stirrup 


‘Ends of Stirrups Looped around 
Tension Reinforcement 






if 
External Shear 


Reaction 
Reaction 


Beam with Vertical Stirrups Beam without Web 
Reinforcement 


Pia. 52.—Action of Vertical Stirrups in Simply Supported Beams. (See p. 152.) 





Ends of Stirrups Looped around 
K Tension Reinforcement 





Column 






Tension Reinforcement. 






(epee eae 
EE CW, 


) le | 
paid EE LA ~Extédynal Shear 


i 
ee of Stirrups Anchored 
in Compression Zone 


Diaganal Crack 


Compression Reinforcement aa 


Fig. 53.—Action of Stirrups in Continuous Beams. (See p. 152.) 


and steel cannot be determined by computation. Assumptions 
variously made are: 


(1) Web reinforcement takes all the diagonal tension with no 
reliance on concrete. The web reinforcement therefore 
resists, in the length s, a total amount of shear equal to 
Vs 
jd’ 

(2) Web reinforcement takes two-thirds of the diagonal 
tension and the concrete the remainder. Web rein- 
forcement resists in the length, s, the force sor 
Where the shearing unit stress does not exceed the 
allowable unit for plain concrete, v’, all stress is taken 
by the concrete. 


154 THEORY OF REINFORCED CONCRETE 


(3) Concrete resists a certain definite unit stress per square 

| inch, v’, the whole length of the beam, and the stirrups 
resist the remainder. Then, in a length, s, the concrete 
resists v’bs, and the web reinforcement resists, 


ae 
zy = 1 — vos = VS — sfos = Pe 

The first assumption corresponds to that made in ordinary beam 
design, where the tensile strength of the concrete is disregarded. 
Tests have shown, however, that the actual stresses in the stirrups 
are less than would be obtained with this assumption. (See p. 40.) 

The second and third assumptions are most commonly used in 
practice. Hither of them will give satisfactory results, especially 
¢ the rule about maximum allowable spacing is followed. The 
Joint Committee on Concrete and Reinforced Concrete, 1916, recom- 
mended the second assumption, while in the later, 1924, reeommenda- 
tions the third assumption is accepted. The authors recommend 
the use of the third assumption. 

Area and Spacing of Vertical Stirrups.—The area of steel and 
the spacing of stirrups may be found by making the force to be 
resisted, as given above, equal to the working strength of the stirrups 
in tension. . 3 

Let V = total vertical shear in pounds at section « feet from left 


support; 

v = total shearing unit stress at section in pounds per square 
inch; 

vy’ = allowable shearing unit stress (or diagonal tension) on 


concrete alone; 

A, = cross-sectional area of all legs of a vertical stirrup in 
square inches. (In a U-stirrup this is the sum of the 
area of the two legs) ; 

fs = allowable unit stress in stirrups in pounds per square 
inch. 

jd = moment arm or distance in inches from center of com- 
pression to center of horizontal reinforcement. (In a 
T-beam, this may be taken as distance between center 

of slab and steel; in a rectangular beam, as 0.87 of the 
total depth to steel) ; 

b = breadth of beam in inches; 

b’ = breadth of web in T-beam in inches; 


I 


DIAGONAL TENSION 155 


$ = spacing of stirrups in inches at a place x Hey from left 
support; 


x = distance in feet from left support to point at which 
required spacing is desired; 
z; = distance in feet from left support to point beyond which 
stirrups are necessary ; 
! = span of beam in feet; 
w = uniform load in pounds per foot. 


Since A, is the area of a stirrup resisting diagonal tension in a dis- 
tance, s, and f, is the tensile strength of steel, the strength of the stir- 
rup in pull is A,f,. The area of stirrups and the spacing for different 
assumptions of distribution of diagonal tension between stirrups and 
concrete may be found as follows: 

(1) Area and spacing if stirrups take all the diagonal tension. 


The diagonal tension to be resisted is a Hence A,f, = “aa nd 


ere _ fajd 
WAS ane ee Ob) and $= “yy Ae oa) 
(2) Area and spacing if stirrups take two-thirds of the diagonal 
tension. 


2Vs 


one 
Diagonal tension to be resisted is 3 id’ Hence A;f; = 3 Gd? and 


we get by solving for As and s, 


2 _ 3f.jd 

As S 3 iad pags (68) HOTS he) 5°) di Rpt ested REED 

(3) Area and spacing if concrete takes a definite amount of shear, v’, 
and the stirrups, the remainder. 


The stress resisted by concrete in the distance, s, equals v’bs. As 





the total stress is Z = “a! the stirrups must carry the difference, 


ase ec, V — v’bjd 
ARR pbSy OF. tegriee 

Equating this to the resistance of the stirrup, A,f,, and solving for 

A, and s, we get 


(V—v jd) . . fsjd 
A, = 3a (70) SCs (V— vijay" : Cub) 


In T-beams, use width of web, b’, in place of b. 


156 THEORY OF REINFORCED CONCRETE 


Area and Spacing of Bent Bars.—The spacing of bent bars, s, 
(or the distance in which they are effective) is usually measured 
along the neutral axis of the beam. The diagonal tension stresses 
to be resisted by the bent bars may be taken, as equal to the stresses 
to be resisted by the stirrups, multiplied by the sine of the angle 
of inclination of the bent bars with the horizontal. Since, however, 
tests indicate that there is practically no difference in effectiveness 
between bars bent at angles between 25 and 50°, it is near enough 
to accept a constant value of 0.7 for the sine for these inclinations. 
Using same derivation as for the stirrups, the formulas for required 
area of bent bars and spacing become as follows: 


(1) Area and spacing if bent bars take all the diagonal tension. 





ae Wee 
A; rae: wg SEED) GS et ye Ma eos 


(2) Area and spacing if bent bars take two-thirds of the diagonal 
tension. 


EN an _ 9 13434 
tinal ee ot (hl O74) nds hae ieee y oe - (75) 


(3) Area and spacing if concrete takes a definite amount of shear, 
»’, and the bent bars, the remainder. 
(V-—v' bjd) . 
fsjd ; 


Uniformly Distributed Loading.—The distance from the support 
to the point where no stirrups are required, for uniform loading, 1s é 


m= 5(1- 5). oy Ral eeu Dae 


Maximum Spacing of Stirrups.—Tests tend to prove not only that 
a sufficient amount of web reinforcement should be provided but also 
that the maximum spacing of stirrups or bent bars should be limited. 


(76) and s = 1.43 | ee ee (77) 


ra (V — v'bjd) 


3The diagram of shearing unit stresses is a triangle from which the distance 


l 
a, may be obtained by the known rule 5 = (5 ae a) my +’. 


This equation solved for a gives formula (76), above. 


DIAGONAL TENSION AIBY. 


If the spacing is too large, cracks may develop between the stirrups. 
A recommendation for maximum spacing is given on p. 250. 

Usefulness of Web Reinforcement.—Numerous tests have 
demonstrated that a beam properly reinforced with stirrups or bent 
bars sustains three or four times as much load as the same beam 
without web reinforcement. The same tests, however, show that 
the web reinforcement retards the appearance of first diagonal cracks 
very little, and that the web reinforcement does not get any stress 
until the first crack appears. It has been noticed * also that under 
working loads (that is, before the diagonal tension exceeds the 
tensile strength of the concrete) the beam acts like a homogeneous 
beam, and as would be expected, the stress in the stirrups is some- 
times compressive instead of tensile. 

This is, nevertheless, no argument against the use of web rein- 
forcement, because in beams without stirrups, final failure follows 
closely the appearance of the first crack, while with beams having 
web reinforcement, stirrups and bent bars represent a factor of 
safety which allows stressing of concrete in diagonal tension nearly 
to its ultimate strength without any danger to the stability of the 
structure. Under working loads the stirrups may not act; but in 
case of overstressing, due to faulty construction or to occasional 
excessive loading, although the concrete cracks, the stirrups not 
only prevent the failure of the beam, but enable it to develop its 
full strength. The minute cracks that may develop are not dan- 
gerous and in many cases are hardly visible. After the excess load 
is removed, the cracks close up. Tests showing conclusively the 
value of stirrups are shown on p. 88. 

Web Reinforcement for Continuous Beams.—The formulas 
given above are based upon results of tests of simply supported 
beams. ‘Their use for continuous beams allows a margin of safety. 

In continuous beams, several conditions tend to prevent, or at 
least to retard, the formation of diagonal cracks. The compres- 
sive force, due to the reaction, tends to close the developed cracks. 
There exists also, almost invariably, some arch action, which de- 
creases the direct. and diagonal tension. In continuous T-beams, 
the horizontal shearing unit stress is zero at the bottom and increases 
till it reaches a maximum at the neutral axis. From there it is 
constant till it reaches the bottom of the flange. As the width of 
the flange is much greater than the width of the stem, the shear- 

4 University of Illinois, Bulletin No. 64, January 13, 1913. 


158 THEORY OF REINFORCED CONCRETE 


ing unit stress in the flange is much smaller than in the stem. Diag- 
onal cracks, therefore, tend to develop in the portion between the 
neutral axis and the bottom of the flange, and larger unit stress is 
required to open them than in simply supported beams. This is 
offset by the fact that at the support of continuous beams, both the 
tensile stresses and the diagonal tension stresses are a maximum. As 
there have been comparatively few tests on continuous beams, the for- 
mulas for web reinforcement given above should be used. 

Web Reinforcement for Cantilevers.—The conditions affecting 
web reinforcement are the same for cantilevers as at the supports 
of continuous beams. In cantilevers supporting vertical loads, 
vertical stirrups must, therefore, be attached to the tension steel 
(at the top) and the free ends hooked in the compressive portion of 
the beam (at the bottom). In cantilevers carrying loads acting in 
other directions, the stirrups must be placed. parallel to the direc- 
tion of the force and attached in the manner suggested above. 


COLUMN FORMULAS 


For reinforced concrete columns centrally loaded, the following 

formulas may be developed: 

Let jf = average compressive unit stress upon the reinforced 
column, equal to the total load divided by the effec- 
tive area; 

f. = compressive unit stress upon the concrete of the column. 
f’.= ultimate strength of concrete cylinders. 

f; = tensile unit stress in spiral, New York Code. 

f’,= compressive unit stress upon the vertical steel in the 


I 


column; 

n= ioe ratio of modulus of elasticity of steel to modulus of 
elasticity of concrete; 

P = load to be sustained by the column; 

A = area of total effective cross section of column (see 


pp. 272 and 406) ; 
, = area of concrete in effective cross section, equal to 
Age Ay, 


A 
A, = area of steel in cross section; 
p = — = ratio of area of steel to total effective area of 


eolumn. | 
’ = ratio of vol f spiral t ] f losed t 
p ratio of volume of spiral to volume of enclosed concrete. 


COLUMN FORMULAS 159 


Since, as is evident from tests, a reinforced concrete column under 
load acts as a unit, the deformation or shortening of steel in the 
column is the same as the deformation or shortening of the concrete. 
stress per square inch 
modulus of elasticity 





From mechanics, = unit deformation, hence 


M4 
a = unit deformation of steel and A = unit deformation of concrete. 
The deformation of steel in a reinforced column is the same as the 


° e E 
deformation of concrete, and since i = n, we have: 
c 


fe_ fe and’ if/s = nf; 
E, EK. s ce 
The stress in steel is therefore equal to the stress in concrete 
multiplied by the ratio of the moduli of elasticity, n. 
If a column sustains a load P, stresses in steel and in concrete must 
be equal to the load. Hence: P = f.A, + PuAPorP: fA) +onfAy 
Since A, = A — A,, we have P = fA(A — As) + nA,l. Finally, 


1: 


ied al VA to 





P=f{A+(n—1A] (79) and f= 


The area of steel, A,, may be expressed in terms of A, by substituting 
A, = pA, which changes the above formulas to 


fr 


Asso) mock, 


P=f.A{1+(n—1)p] (81) and iis 
Knowing the stress, f,, and the steel ratio, p, we may find the 


required area from 
P 


— fdl+(n—Wpl * 


Knowing the stress, f., and the total area, we may find the required 
area of steel, A,, and the percentage from 


A (83) 


pe DE RS vah 


(85) 


The average unit stress, which is the total force, P, divided by the 
effective area, A, 


‘2 Pe 
ai, and Nie? 


160 THEORY OF REINFORCED CONCRETE 


The relation between f and f, may be found by substituting for P 
in the above equation its value from Formula (81), giving 


f= fii + (n— Vol. OS eee 


Values of f for different percentages of steel are given on p. 916. 

Columns with Spiral Reinforcement.—The ultimate strength of a 
column with spiral reinforcement depends upon (1) the amount of 
vertical steel, and (2) the amount of spirals. Therefore, in formulas 
for the breaking strength of a spiral column, the amount of spirals 
must be considered. In design, however, the elastic limit and 
not the breaking strength of the column is the determining value, 
as explained on p. 82. } 

There are differences of opinion as to the proper formulas for 
spiral columns. Practical design is treated on p. A419. 

The following formulas are in use: 

(1) Columns with spiral limited to 1 per cent of the volume of 
concrete within the spiral, and variable amount of vertical steel. 

Same formulas are used as for columns with vertical steel only. 
To allow for the effect of the spiral, the unit stresses in concrete and 
vertical steel are increased. | 

This method is specified by the Codes of Boston, Cleveland, and, 
with some modification, Philadelphia. It was recommended by the 
Joint Committee in 1916. Although it does not fully represent the 
actual conditions, the authors consider this method as the most rea- 
sonable thus far developed. ; 

(2) Columns with spiral varying from one-half to 2 per cent of 
the volume of concrete (p’ = 0.005 to 0.02) and variable amount of 
vertical steel. 

Formula for strength of column (Building Code of New York). 


ig aa Arie me Te ele se 27 Ae 


In this formula, the stress in concrete and vertical steel is assumed 
to be the same as for columns with vertical steel only. The spiral is 
assumed to increase the strength of the column by its strength in 
tension. (Any section cuts the spiral at two places, which explains 
the value 2 in the term 2p'A fs.) 

(3) Columns with spiral from 3 to 14 per cent of the core (p’ = 
0.005 to 0.015) and variable percentage of vertical steel, but not to 
exceed the percentage of spirals. 


COLUMN FORMULAS 161 


Formula for strength of column (Building Code of Chicago). 
Seda Wt 1p + 28(n — yp]. (87) 


In this formula, the effect of spiral is considered as two and one- 
half times as large as the effect of same amount of vertical reinforce- 
ment. In addition, the allowable stress, f., is larger than for columns 
with vertical bars only. 

(4) Joint Committee 1924 formula. Percentage of spiral shall 
not be less than one-fourth the percentage of vertical steel. Vertical] 
steel shall be from 1 to 6 per cent of the core. 

The formula for strength of column is of the same form as that 
for columns with vertical steel only; the difference is in the allowable 
unit stress in concrete, which is a function of the ultimate strength 
and of the steel ratio, :. 


Pew a hi let (i alpen (88) 
~ Where the allowable stress is 
fee 200 + (O10 anyfie Oy (89) 


The value of f’. is the ultimate compressive strength of concrete 
at age of twenty-eight days, based on 6 by 12 in. or 8 by 16 in. eyl- 
inders. 

In this formula, an attempt is made to take into account the 
stresses in steel due to shrinkage and yield of concrete, in accordance 
with the ideas advanced by Mr. Franklin R. McMillan in his paper, 
“Study of Column Test Data.’ 5 

The authors consider that the refinements and the change from 
present practice required by these new formulas are not yet war- 
ranted by our present knowledge of the subject. The formulas are 
not well adapted for use without tables, because a change in per- 
centage of steel not only affects the amount of total load carried by 
steel, but also the unit stress in concrete. For instance, to find the 
amount of steel for an accepted size of column (a very common 
problem) requires solution of two simultaneous equations, one of 
which is a second-degree equation. Moreover, the cost of columns 
required by these formulas is larger than by the formula of the 
previous Joint Committee recommendation, while there is not yet 
sufficient proof that the previous formula did not give safe results 
to warrant the increase in cost of construction. 

° Proceedings, American Concrete Institute. Vol. XVII, 1921, p. 150. 


162 THEORY OF REINFORCED CON CRETE 


MEMBERS SUBJECTED TO DIRECT TENSION 


In concrete construction, there very often occur members sub- 
jected to direct tension, as for instance, stair hangers, floor hangers 
in arch bridges, tie-rods in arches, walls in circular tanks, conical 
hoppers in bins. Such members are subjected to direct tension only. 
Any bending in the direction of the tensile stress is only incidental. 

Members like the bottom chord in concrete trusses, particularly 
when the truss is without diagonals as in Vierendeel trusses, are 
subjected to tension and a bending moment. The same is true of 
the sides of square or rectangular reservoirs. 

The stress conditions are discussed below. 

Members under Direct Tension Only.—When a member is sub- 
jected to direct tension only, all stresses must be resisted by steel 
and no allowance should be made for the strength of concrete. 

The required area of steel is 


(90) 


where T is the total tensile stress and f; the allowable unit stress in 
steel. 

The unit stress, fs, for which the reinforcement is to be designed, 
depends upon conditions. Where the stretch due to the stress is 
not harmful, a stress in steel of 16.000 Ib. per sq. in. may be used. 
In circular tanks designed to contain fluids, the stresses in the rings 
must be limited to from 10000 to 12 000 lb. per sq. in. to limit the 
stretch as much as possible and thus avoid open cracks in concrete 
and the consequent leakage. Also, deformed bars are recommended 
to properly distribute any cracks. The hair cracks which may occur 
for these recommended stresses are not likely to extend the full 
thickness of the wall and therefore would not affect its water-tight- 
ness. | 

In designing arches with tie-bars, their stretch must be considered 
(See Arch Design, Volume III.) 

Composition of Tension Members.—When a tension member is 
not likely to be subjected to secondary bending, it may consist of a 
single bar. When imbedded in concrete as a protection against rust 
and fire, the bar is placed in the center of the concrete section. 

In a tension member likely to be subjected to secondary bending, 
as, for instance, the hanger for floor beams in arches which may be 


MEMBERS SUBJECTED TO DIRECT TENSION 163 


struck by a vehicle, the reinforcement should consist of four bars 
placed in the corners of the concrete section. 'The member is then 
able to resist any such bending moment. Since the stresses due to 
bending moment increase the tensile stresses due to direct tension, 
it is necessary to use a larger area of steel than would be required 
_ for direct tension alone. Sometimes the bars for direct tension are 
placed in the center of the member and small bars are added in the 
corners of the concrete section for bending. 

Details of Tension Members.—It is important to keep the 
tension members straight. It is obvious that a member with an 
accidental curvature would straighten under tension, and the move- 
ment due to this straightening would be injurious to the rest of the 
construction. A sag in horizontal ties also should be avoided. 
For these reasons, in arches, the tie is provided with a turnbuckle 
for the purpose of tightening it after installation. The tie-bar is also 
supported by suspenders at proper intervals so as to prevent sagging 
under its weight. 

If the tension member consists of a number of bars, they should 
be held properly in place by ties or hoops in the same manner as 
column reinforcement. 

The bars composing the tensile reinforeement must be anchored at 
the ends, so as to develop their full strength. The anchorage may 
consist of straight imbedment, of straight imbedment with a hook, 
or of a screw thread at the end with a nut and a plate. In circular 
tanks, the rings are usually spliced by lapping. It is important that 
the splices of the rings be staggered. 

With straight imbedment only, the bar must be extended into 
the anchoring member a sufficient distance to develop by bond the 
strength of the bar. (See p. 268.) 

_ With straight imbedment and hook, the straight imbedment 
may be made shorter because of the effect of the hook. In important 
members, the straight imbedment is made long enough to satisfy 
bond requirement and the hook is added as an additional factor of 
safety. The bearing stresses of the hook in the concrete are high, 
although with properly designed shape of hook (see p. 269) they are 
not excessive. ‘To relieve the stresses, the hook may engage a cross 
bar; this distributes the compression stresses and also tends to pre- 
vent any tendency to splitting. The plate method is illustrated in 
Fig. 218, p. 679. In this method the bars are threaded at the end, 
a plate of proper dimensions is placed at the end, and this is kept 


164 THEORY OF REINFORCED CONCRETE 


in place by nuts, one for each bar in front, and the other back of the 
plate. When ‘mbedded in concrete, the plate distributes the tensile 
stresses on the concrete. ‘The area of the plate is determined by 
dividing the total tension in the bar by the allowable bearing stress 
on concrete. If more than one bar is used in the member, each bar 
may be provided with a separate plate or one plate may be used for. 
all bars. The thickness of the plate should be determined by con- 
sidering it supported by the bars and loaded uniformly over its: whole 
surface by the reaction of the concrete. When a plate is used for a 
single bar, it should be treated as a cantilever. When one plate is 
used for two bars, it may be considered as a simply supported plate 
uniformly loaded. The plate must be imbedded far enough beyond 
the end of the tensile member to prevent shearing of the concrete. 

Thickness of Concrete in Tension Members.—In fireproof con- 
struction, the tensile members should be imbedded in concrete to 
protect them from fire and rust. If subjected to tension only, the 
required concrete section is governed mainly by the required thick- 
ness of the protective cover. 

Where lateral stiffness of the member is desirable on account of 
the possibility of accidental stresses, the dimensions should be 
adopted accordingly. In circular tanks, the thickness of concrete 
should be sufficient to make them watertight. 


MEMBERS UNDER FLEXURE AND DIRECT STRESS 


The formulas given below apply when any member of a structure 
is subjected to compression (or tension) due to a central force and 
to bending or flexure. This takes place when the member is sub- 
jected: | 

(1) Simultaneously to a bending moment and a central force or 
thrust; 

(2) To an eccentrically applied force or thrust. 

The first condition occurs in columns carrying the column load 
and in addition—particularly in wall columns—subjected to a bend- 
ing moment caused by rigid connection between beam and column. 

Another instance is that of columns with brackets for crane 
runways. The crane load produces a bending moment in the col- 
umn in addition to the central force. All members in rigid frames, 
particularly the vertical members, are subjected to central force 
and bending. In trusses, the compression members must resist, 
in addition to the central force, a bending moment, which in trusses 


DIRECT STRESSES AND FLEXURE 165 


with diagonals is produced by the rigidity of connection at the joints, 
and in trusses without diagonals by the shear, as explained in Volume 
III of this treatise. 

The second condition occurs in arches and dams when the line 
of pressure intersects the normal section at a distance from the arch 
axis, in other words, when the thrust is applied eccentrically. The 
thrust may be normal to the section or, as is usually the case, it may 
be inclined to the section. Before determining the stresses, the 
inclined thrust, 2, is resolved into a component normal to the sec- 
tion, V, and a component parallel to the section, V. The normal 
component produces both direct stresses on the section and bending, 
while the parallel component produces only shearing stresses which 
are usually small and do not need to be considered in design. 

An eccentrically applied force is found also at the base of retaining 
walls where the resultant pressure is composed of the horizontal earth 
pressure and the vertical weight of the structure. In this case, the 
formulas below are used to find the distribution of pressure over the 
foundation. | 

Sign of the Thrust.—The thrust is either positive, when it pro- 
_ duces compressive stresses on the section, or negative, when it pro- 
duces tensile stresses. Usually, the positive thrust is called “ thrust,” 
and the negative thrust, ‘ pull,’”’ although the desiginatons “‘ positive 
and negative thrust ” are often used. 

Relation between Bending Moment and Eccentrically Applied 
Thrust.—The stresses produced by a combination of a central 
thrust and a bending moment are the same as those produced by an 
eccentrically applied thrust. Thus, a central thrust, V, and a bend- 
ing moment, M, may be replaced by an eccentric thrust, N, acting 
at a distance from the axis of the section equal to e = NV" In turn, 
the eccentric thrust may be replaced by a central load of the same 


intensity and a bending moment equal to the thrust multiplied by 
the eccentricity.° Therefore, both cases can be solved by the same 


6 Proof of the above statement. Assume that a section is submitted to an 
eccentric thrust N acting at a distance e from the axis. The stress conditions will 
not be altered if in the center of the section are added two equal forces, but acting 
in opposite directions, such as + N and ~ N. The section is then exposed to 
three forces, namely + N at a distance e from center of section and — N and + N 
in the center of the section. The eccentric force with the negative central force 
forms a couple, Ne, and may be replaced by a bending moment M = Ne. ‘There 
remains in the center a positive force N. Thus the eccentric thrust is replaced 
by a central force, N, and a bending moment M = Ne. 


166 THEORY OF REINFORCED CONCRETE 


formulas. The case of an eccentrically applied thrust gives simpler 
formulas. 

Relation between Position of Eccentric Thrust and the Sign of 
Bending Moment.—A _ positive bending moment, producing in a 
horizontal member,’ compressive stresses at the top of the section 
and tensile stresses at the bottom, may be replaced by a positive 
thrust acting above the axis. 

A negative bending moment, producing compressive stresses 
at the bottom of the section and tensile stresses on the top, may 
be replaced by a positive thrust acting below the axis. 












Gravity Axis __ _ Gravity Axis P 


Eccentric Thrust applied cnut : Central Thrust and 
above Gravity Axis- puivniest te Positive Bending Moment 


Fic. 54.—Positive Bending Moment and Central Thrust. (See p. 166.) 







_ Gravity Axis 4 





Eccentric Thrust applied Ea enee Central Thrust and 
below Gravity Axis Shahan ae ne Negative Bending Moment 


Fic. 55.—Negative Bending Moment and Central Thrust. (See p. 166.) 


Conversely, a positive thrust acting above the axis produces 
maximum compression at the top and minimum stresses at the bot- 
tom. A positive thrust acting below the axis produces reverse 
results. 

Negative Thrust, or Pull—tIn the discussion, that follows N is 
assumed as a positive thrust, ie., a thrust producing compression, 
and therefore the maximum stresses are compressive. If N is a 
negative thrust, or pull, as in arch design with the thrust due to 


7 Similar action, of course, is true in a vertical or inclined member. 


DIRECT STRESSES AND FLEXURE 167 


rib shortening or to fall of temperature, it will produce tension over 
the section. A negative thrust, applied below the gravity axis, 
produces positive bending moment, while, if applied above the AXIS, 
it produces a negative bending moment. Conversely, positive 
bending moment and pull can be replaced by an eccentric pull acting 
below the gravity axis. The reverse is true of negative bending 
moment and pull. 

Negative thrust (or pull) combined with positive bending moment 
(or negative thrust acting below axis) produces maximum tension 
at extreme fiber below the axis and minimum tension (or small com- 
pression) at the extreme fiber above the axis. Pull combined with 
negative bending moment (or negative thrust acting above the axis) 
produces reverse results. The formulas given in subsequent pages 
may be used for negative thrust (or pull) by substituting —N for N. 
This changes the compression unit stresses due to the thrust to 
tensile unit stresses and reverses the position of bending stress in the 
section. 

A question arises as to whether, with negative thrust, it is per- 
missible to use the formulas given below for materials such as con- 
crete, which is not capable of resisting tensile stresses. If the nega- 
tive thrust and bending moment act alone, then, of course, the 
formulas are not applicable and the formulas on p. 189 to 194 should 
be used. When, however, the tensile stresses due to the negative 
thrust are counteracted to a great extent by a positive thrust, the 
formulas in this section, after the sign is changed before N, are appli- 
cable. For instance, in arch design, the negative thrust due to fall of 
temperature is counteracted by the positive thrust due to dead load. 
In such a case, the stresses due to both thrusts are computed by the 
same formulas separately and the results added. 


NOTATION 


Let #& = inclined thrust acting on a section; 
N = normal thrust, a component of thrust R normal to the 
section; also a central load; 

V = external shear, the component of 2 parallel to the sec- 
tion; 

T = normal pull, in members subjected to direct tension or 
tension and bending; 

b = breadth of rectangular cross section; 


168 


€0 


itn 


THEORY OF REINFORCED CONCRETE 


height of rectangular cross section; 

eccentricity, that is, the distance from axis of gravity of 
the section to the point of application of the thrust; 

value of eccentricity which produces zero stress in con- 
crete at outer edge of rectangular section; 

bending moment on the section; 

perpendicular distance to any point in the section from 
eravity axis of section; 

moment of inertia of cross section of concrete about 
the gravity axis; 

moment of inertia of cross section of steel about the 

horizontal gravity axis; 

moment of inertia of transformed section; 

total effective area of cross section of concrete; 

total area of section of steel; — 

area of transformed reinforced concrete section; 

perpendicular distance from outside fiber to gravity 
axis of unsymmetrical section having maximum 
compression ; 

perpendicular distance from center of gravity of unsym- 
metrical section to outside fiber having maximum 
tension or minimum compression ; 

maximum unit compression in concrete; 

maximum unit tension in concrete, or minimum com- 
pression ; ate 

maximum unit compression in steel; 

maximum unit tension or minimum unit compression 
in steel; 

ratio of steel to total area of section; 


es ~ yatio of moduli of elasticity of steel and concrete; 


ratio of depth of neutral axis to depth d; 

distance from outside compressive surface to neutral 
AXIS; 

depth of steel in compression; 

depth of steel in tension; 

distance from center of gravity of symmetrical section 
to steel; 

constants. 


PLAIN CONCRETE UNDER COMBINED STRESSES 169 


PLAIN CONCRETE SECTION UNDER DIRECT STRESS AND BENDING 
MOMENT 


General Formula.—For members of any cross section subjected 
to a central load, N, and a bending moment (or to an eccentric thrust), 
the stresses may be obtained by computing separately the stresses 
caused by the central load and by the bending moment. The sum 
of the results then gives the actual stresses. 

The unit stresses produced by a central load, N, obtained by 
dividing the load by the effective area of the section, are equal to _ 

The unit stresses produced by the bending moment, M, at any 
point at a distance, y, from gravity axis, from simple mechanics, 
are equal to + ee The sign + is used for points above the axis, 
and — for points below the axis. 

The combined stresses, therefore, at any distance, y, from the 
axis of gravity, are equal to the sum of the two above expressions. 

Stresses at any point. 

NM 

io A= ie (91) 

The second term of this expression, depending upon the value y, 

varies with the position of the point in relation to the axis of gravity, 

therefore, the intensity of stresses varies from a maximum at one 
edge of the section to a minimum at the opposite edge. 








Gravity 
Axis 





aie 
Section Stresses due to Stresses due to Combined 
, Bending Moment. M. Central Thrust, N. Stresses 


Fig. 56.—Section Subjected to Direct Stress and Bending. (See p. 169.) 


Usually, only the maximum and minimum stresses in the extreme 
fibers are of interest. If, as given in Fig. 56, y: is the distance of 


170 THEORY OF REINFORCED CONCRETE 


the extreme fiber above the axis and y2 below the axis, the combined 
stresses are: : 


Maximum Stresses, 


= N My 
he aE. A + yi ° . e ° ° e » e (92) 
Minimum Stresses, 
oe N oy Mype 


This condition exists for a positive bending moment, or when 
the eccentric thrust is applied above the axis. 

For negative bending moment or a thrust acting below the axis, 
the sign before the second term in the formulas changes. The 
maximum stress acts below the axis and the minimum stress above 
the axis. 

Formulas for Rectangular Plain Sections.—In a rectangular sec- 
tion with breadth, b, and height, h, the area A = bh, the moment of 


inertia [ = te and the distance from axis to extreme fibers, 


UIE AS These values substituted in the general equation 


give the following expressions for rectangular section: 


Maximum Compressive Stress, 


N , 6M 
Se a bh + Bh2’ = F 3 ° . : 5 (94) 
Minimum Stress, 
N 6M 
fi = pp — Bye (95) 


If the bending moment M is replaced by Ne the above formulas 
change to: 


Maximum Compressive Stress, 


Minimum Stress, 


f= (+7). We 


h 
hisae (1 . i) (97) 


PLAIN CONCRETE UNDER COMBINED STRESSES 171 


For positive bending moment (or thrust applied above the axis), 
maximum stresses act above the axis. The position of maximum 
stresses is reversed with negative bending moment or with thrust 
below the axis. When 0, fh and e are in inches and M isin inch-pounds, 
and N in pounds, the unit stresses are in pounds per square inch. 
For M in foot-pounds, multiply the second term by 12. 

To get the stresses in pounds per square foot, as is customary 
in dam design, the dimensions b and h must be in feet and M in foot- 
pounds. | 





Force at Edge of the 
Middle Third. Middle Third. 


Fic. 57.—Stresses Caused by Eccentrically Applied Thrust. (See p. 171.) 






£4 
et 
S| 


eth 
3 





es 





Via. 58.—Stresses Caused by a Force Acting Outside the Middle Third of Plain 
. | Concrete Section. (See p. 172.) 


For positive thrust, the maximum stress is always compression. 


The minimum stress from Formula (97) is compression when - is 


_ smaller than unity, that is, when the eccentricity, e, is smaller than 
: When the force acts at the edge of the middle third of the section, 
the minimum stress equals zero, and the maximum stress equals 
double the stress caused by a central load of equal intensity. (See 
mie. of, p. 171.) 


172 THEORY OF REINFORCED CONCRETE 


When e¢ is larger than i and the load acts outside of the middle 


third, the minimum stress is negative, and the section is subjected to 
tension. In such cases, this formula can be applied only when the 
material is capable of resisting tensile stresses. 

If the material cannot resist tension, as in masonry, Or in con- 
crete where tension exceeds the allowable stress, it is necessary to 
assume that the pressure is distributed only over a section equal to 
three times the distance of the point of application of the load from 
the nearest edge. (Sce Fig. 58). If that distance is g, the total 
effective depth of the section is 39. Substituting in Formula (96), 


e = | andh = 39, we get for plain concrete and masonry. 


Maximum Compression, 


ate a 3b an ble SRR che ae eens (98) 


Plain Circular Sections.—lor circular sections where d is diameter 


4 
of circle, the area of moment of inertia is J = ie and the section 


is A = ae — (,785d?, and the distance to extreme fiber yi = Yy2 = = 


These values substituted in equations 92 and 93 give 


Maximum Compression in Concrete, 


N 10.2M (99) 








Se = D751 7 de 

Minimum Stress, 
N 10.2M 

i= Tye (109) 
Substituting 

M = Ne. 
Maximum Compression Stress, 

ne zp( 1.275 ab 0). i 
Minimum Stress, . 

fz (1-275 a 02). ON eee 


The stresses at one extreme edge become zero for e = 0.125d. 
For larger eccentricity, tensile stresses will be developed. 


DIRECT STRESSES AND FLEXURE 173 


REINFORCED CONCRETE SECTION SUBJECTED TO DIRECT STRESS 
AND FLEXURE 


General Formulas.—The stresses in a reinforced concrete section 
subjected to direct stress and flexure may be obtained by replacing 
the reinforcement by an equivalent area of concrete and then apply- 
ing to the transformed section the general Formulas (92) and (93) 
developed for plain concrete sections. To get the equivalent area of 
concrete, the area of reinforcement is multiplied by the ratio of moduli 
of elasticity of steel and concrete, n. The area of the transformed 
section is A: = A+ (n—1)As. The moment of inertia of the 
transformed section about the axis of gravity equals the moment 
of inertia of the concrete section plus the moment of inertia of the 
concrete replacing the steel. The substitute area must be placed 
at the same distance from the axis as the steel area, so that the 
moment of inertia of the substitute area equals (n — 1) times the 
moment of inertia of the steel area about the gravity axis of the sec- 
tion. The total moment of inertia, I, =I +(n—1)I;. The 
value of (n — 1) is used in the formulas instead of n because, in taking 
the concrete area, one A is already included. Substituting these 
values in Formulas (92) and (93), the formulas for maximum and 
minimum stress become 


Maximum Compressive Stress, 


N My 








Peas, PTE Gh) 40) 
Minimum Stress, 
| my, N Mye 
Meena =A, Ton DL (104) 


The stress in steel equals the stress in the adjoining concrete 
multiplied by n. As evident from the formulas, for positive bend- 
ing moment, maximum stresses act above the neutral axis while 
minimum stresses act below the neutral axis. For negative bend- 
ing moment, the sign before the second term of the equation changes, 
and the maximum stresses act below the neutral axis while minimum 
stresses act above the neutral axis. | 


174 THEORY OF REINFORCED CONCRETE 


RECTANGULAR SECTION WITH SYMMETRICAL REINFORCEMENT 


The formulas for a rectangular section subjected to direct load 
and bending moment depend upon the arrangement of reinforcement. 
The bars may be placed in the section as shown in Fig. 59(a) or as 
shown in Fig. 59(0). 


_-Bars f, 
Per, 








*<t gars g 
Gravity Axis ag 


Transformed Section (a) Transformed Section (b) 


Fra. 59.—Arrangement of Bars in Rectangular Section. (See p. 174.) 


In Fig. 59 (a), the area of concrete is A = bh, the area of steel, 
be 


I 


pbh, and the transformed area, 


A, =A+t(n— DA, = dll + (nm — Vol 


Il 


pet bh? 
The moment of inertia of the concrete section is I. = 73° Since 


all the bars are at the same distance from the gravity axis, their 
moment of inertia is obtained by multiplying their area by the 
square of the distance, a. The moment of inertia of steel is I, = Aue 
and the moment of inertia of concrete replacing the steel is 
(n — 1)A,a?. The total moment of inertia is therefore 


DIRECT STRESSES AND FLEXURE 175 


3 
I=I1,+ (n— 1)A,a? = a + (n — 1)pbha? 


pe 2 J 
= On T+ (n= type. ieee ae eee) (LOD) 


The value of (n — 1) is used in the formulas instead of n, because 
in taking bh for concrete area one A, is already included. 

For the arrangement of steel shown in Fig. 59 (6), the trans- 
formed area may be found in the same manner as before. It is 
A, = bh[1 + (n — 1)p]. The formula for the moment of inertia 
of concrete is the same as before, but the formula for the moment of 
inertia of steel is different because the distance of the side bars, g, 
from the gravity axis is different from the distance of bars Siege 
formula taking the bars, g, into consideration would be too com- 
plicated. Since their effect on bending stresses is comparatively 
small, they may be disregarded and the same formulas used as 
for case (a). The value of p to be used in the formulas should be 
based on bars f only, or else a small allowance may be made for the 
effect of bars, g, by using, in figuring p, an area of steel somewhat 
larger than in bars f. 

Thus, in both cases the transformed area is bh[1 + (n — 1)p] 


and the moment of inertia J, = on| + (n — 1)pa?). Substi- 
tuting these values in the general Formula (91) and making M = Ne, 
we get 


Stresses at any distance, y, from gravily axis 


ay’ il 12ye : 
Je = ae + (n= l)p a h? + 12(n — Saat mG 


The maximum and minimum stresses in concrete are at the edges 








of the section for y = h 











5 
Maximum Compression Stress in Concrete, 
N 1 6he 
Je = sali +) (ni—1)p ae h? + 12(n — Twat: mAh 
Minimum Stress in Concrete, 
N 1 6he 
Die aA +(n—1l)p h+12(n— naw el ihsd 


The stresses in steel equal n times the stresses in concrete at a 
distance a from the axis. Thus, 


176 THEORY OF REINFORCED CONCRETE 


Maximum Compression Stress in Steel, 


N 1 12ae 
f= "slit @oie | Pt ee ope eth) 


In the limits within which these formulas are applicable, the 
minimum stress in steel has no significance and does not need to be 
computed. 

In the above formulas, if N is in pounds and }, h, and e in inches, 
the resulting stress is in pounds per square inch. The value e is 
obtained by dividing bending moment, M, in inch-pounds, by the 
thrust, N, in pounds. 

To get stresses In pounds per square foot, the values ), h, a, and ‘e 
must be in feet. The value of ¢ is obtained by dividing M in foot- 
pounds by the thrust in pounds. 

For positive bending moment, the eccentric thrust is applied 
above the axis and the maximum compression stresses are at the 
upper edge of the section. 

For negative bending moment, the eccentric thrust is applied 
below the axis and the maximum compression stresses are at the 
lower edge of the section. 

The resulting stresses are shown in Fig. 60, p. 176. 





1 : 
= pbhf,-- 
Force Producing Compression upon the Force Acting at a Distance 
Whole Reinforced Section. . Larger than eo from the Axis 
of Gravity of Reinforced 
Section. 


Fic. 60.—Stresses in Reinforced Concrete Section Subjected to Eccentrically 
Applied Thrust. (See p. 176.) 


Solving Formula (107) by Diagram.—In Formula (107), the 
xpression C. = (s—-7 9, T _ i ee 
expression Ce= \7 7 (m— Ip | hk? + 12(n — Vpar 


for constant values of p, ; and -. The formula changes to 


) is a constant 


DIRECT STRESSES AND FLEXURE 177 


Maximum Compression Stress in Concrete, 


N 
ee PETS Peppered Buk rie) G10) 


The value of C, is given in diagrams on p. 935. Separate dia- 
grams are drawn for a = 0.5h, a = 0.45h, a = 0.4h and a = 0.35h. 


The variables in each diagram are : and p. 


For the purpose of making a diagram the constant is written in 
the following form: 


1 6 


€ 
2 h (111) 
~ T+ (@— typ ehinanie 


C, ——____ 
1 + 12(n ~ 1)p(¢) 


The use of the diagrams may be seen from the example below. 


Example 1.—Determine stresses in concrete column when 6 = 12in., h = 24 
in.; the reinforcement consists of six g-in. round bars; fireproofing 2 in. and the 
section is subjected to a normal thrust, VN = 550 000 lb. and a bending moment 
M = 2700 000 in.-lb. Consider total area as effective. 

Solution.—The area of concrete is bh = 12 X 24 = 288 sq. in. Since the 


area of a j-in. round bar is 0.6 Sq. in., the area of steel is 6 X 0.6 = 3.6 Sq. in. 


. 3.6 
The steel ratio, p = —— = 0.0125. Since the fireproofing to outside face of bars 
. 288 


is 2 in. and the bar is { in., the distance to center of bar, d’, is 2 + 7% = 2% in. 


9-2. 
The value of a isa = 24 — 276 = 9% in. The ratio ; is ai = 0.40. 
M 2 700 000 
The eccentricity, e, found from W is ¢ = 550000 7 5 in. and the ratio 
5 

; ae 0.208. The value of C, is now found for a = 0.4 h, p = 0.0125 and 
ers a” 
h aad . . 


Use the diagram marked 24 = 0.8h. Find the intersection of a vertical line 
e 
corresponding to bak 0.208 with a curve for p = 0.0125 (interpolate between 


p = 0.012 and p = 0.014). Read at the left the value of C, = 1.79. 
The maximum compression then is 


550 000 ; 
fe =1.79X 12x24 = 342 lb. per sq. in. 


Effect of Eccentricity.— As in plain concrete sections, the location 
of the center of thrust determines the distribution of the stress. If 


178 THEORY OF REINFORCED CONCRETE 


the thrust acts at the center of gravity, there is uniform compression 
over the whole section. As the center of thrust moves from the 
pravity axis, the compression at the opposite surface decreases until 
it finally becomes zero, and then changes to tension. 

When the first term in the brackets of the above equation is 
oreater than the second, the minimum stress in the concrete 1s com- 
pression. When the two terms are equal, the stress is zero at the 
outer edge of the concrete on the opposite side to that on which the 
thrust acts. When the second term is greater than the first the result 
is negative and the minimum stress is tension. 

If the tension, determined by the above formula, exceeds the 
allowable tension on concrete, the above formulas are not applicable 
and the formulas given on p. 180 should be employed. 

Thrust Applied so that the Compression at One Surface Becomes 
Zero.—The eccentricity for which this occurs may be determined 
by equating the two terms in Formula (108) and solving the resulting 


equation for e. 



















| 

J 
ZIR 

| 

| 


on i ay Gravity Axis, 
\ 








A pbhf,— D 
Bird al 


Fic. 61.—Stresses Caused by a Force Acting at a Distance equal to ¢ from 
the Axis of Gravity of Reinforced Section. (See p. 178.) 


Using previous notation and also letting eo = value of e, for which 
the minimum stress is zero, then 


Limiting Eccentricity, 


++ 12(n — ve(§) 


eo - —gi+@— Dpl er 





In the above case, the formulas on p. 175 change to 


DIRECT STRESSES AND FLEXURE 179 


Maximum Unit Compression in Concrete, 


2N 


eee Gs Spor Pl ere eo eee oe ay tn ae (113) 


Maximum Unit Compression in Steel, 





ve nN 2a 

1 ibe acess aay aeeeteee 
also 

é eae 

fs = fo. e ® ° e ® ° e ° (115) 


Minimum Compression in Concrete = 0. 
_ Minimum compression in steel is very small and does not need to 
be determined. 
In the diagrams on pp. 935 to 938, the dash curve indicates the 
e 
h 
zero. If it is desirable to avoid tensile stresses, the dimensions of the 


limiting values of = for which compression at one surface becomes 


section should be so selected that the : is smaller than, or equal to, 
eae €o 
the limiting value he 


Effective Section.—Some building codes require that the outside 
protective cover of concrete be disregarded in computing combined 
stresses, in the same manner as recommended for centrally loaded 
columns. Then the values of b and h in the formulas are the dimen- 
sions within the fireproofing. For instance, if 13 in. fireproofing is 
required, a 12-in. by 18-in. section becomes 9 in. by 15 in., because 
Meee X11.5)'= 9 and (18 — 2 x Toya 153 

Circular Reinforced Section.—In a circular section, the reinforce- 
Ment is evenly distributed along the circumference. It may be 
replaced by a ring. The area of the section equals the area of con- 
crete, A, plus (n — 1)A. The area of the circle is 0.78542. If p 
is the ratio of steel, 

A, = 0.785d?[1 + (n — 1)p]. 

The moment of inertia of the concrete section, I, = ee The 
moment of inertia of the steel ring the area of which is A, and the 
A,a? 


radius, a, is I, = a 





180 THEORY OF REINFORCED CONCRETE 


Since A, = pA = 0.785pd?, the moment of inertia becomes 
2 
I, = pas = 0.393pd2a2._ The total moment of inertia, of trans- 
formed section, 
ad* 
I,= 64 +(n- 1)0.393 pd?a? 
Finally 
L= dt| = + 0.393(n — no 
20 4 Foe d 
The values of A; and J, substituted in Equations (103) and (104), give 


Maximum Compression Stress, Circular Reinforced Section, 


N M 





Den) aie ; J ee eee (116) 
0.785d2{1 + (n — 1)p) eluinee _fa\? 
od Fee: 0.393(n 1) (5) | 
Minimum Stress, Circular Reinforced Section, 
S 1 ae Sy) fa 





fe oe ae ee eee 
0.785d2{1 + (n — 1)pI lea y a\* 
2d ona 0.393(n (5) 


Distribution of Stress in Rectangular Section When One Surface 
is in Tension.—When the thrust is applied at a distance from the 
gravity axis greater than the eccentricity, eo (from Formula (112), 
». 178), and the concrete is assumed to be unable to carry any ten- 
sion, then Formulas (107) and (108), p. 175, are no longer applicable. 
The following method may be used in determining the conditions 
ander which the steel on the side opposite the thrust is assumed to 
resist all tension stresses. Referring to Fig. 62, p. 181, and making 
the same assumptions as given in connection with simple flexure on 
p. 126, the following relation is found between the stresses 1n steel 
and in concrete. If fe is compression in extreme fiber, then 


Unit compressive stress in the upper steel 18, 


d’ 
f= nfe(l - ia) AMR re: aaa Ri a (118) 
and : 
Unit tension in the lower steel is, 


d — kd 1-k 
Te = nfo = Oe ee da ee (119) 





DIRECT STRESSES AND FLEXURE 181 


The magnitude of the stress, f., may be determined from the principle 
that for equilibrium the sum of the stresses acting on a section must 
equal the thrust, and the principle that the bending moment of the 
external forces (which is the thrust multiplied by the eccentricity) 
equals the moment of resistance of the internal stresses, 

In this case, it is more rational to use the depth to steel, d, as a 
basis instead of the total depth, h, as used in previous formulas. 


A, : 
The steel ratio, p, equals bd where A, is the total area of steel in the 


section. For symmetrical arrangement of bars, the areas of tensile 
and compressive reinforcement are equal to each other and amount to 
3pbd. 

gba, 
















Sf Sys Gravity Axis} 
opt Baan re career: fase 
_Y¥ H ' Piles Neutral Axis | 








Fig. 62.—Stresses Caused by a Force Producing Compression and Tension 
upon a Reinforced Section, Tensile Strength of Concrete Neglected. (See 
p. 180.) 


From the first principles, we have 


_ fd , fdkd _ fopbd 
eee Wu Ga 


Substituting the values for f’, and f, from formulas (118) and (119), 


h 
2 ei eae 
Pe k* -- 2npk Npa 


Ne i as 


(121) 








The moment of the stresses about the gravity axis, obtained by 
taking the sum of the moments of all the stresses about the gravity 
axis, after eliminating f’, and f, by the use of Equations (118) and 


(119), is , ee ede 
eyes (neye ty Sey aes 


182 THEORY OF REINFORCED CONCRETE 


By equating the expressions in Formulas (121) and (122) to the 
known thrust and known bending moment, we get two equations 
from which the unknown values of k and f, may be determined. 
This would mean, however, solving third power equations. In 
practice, the use of the curves given on pp. 938 and 939 will be found 
convenient. 

Calling the quantity in brackets in Formula (122) 


C, = [2(5) + aa) -F | og ies ae 


Cofcb0?, 06s) ) (124) 


we may write 


I 


from which 
Maximum Compression Stresses in Concrete, 


M 
age Cbd’ (125) 
and 
Maximum Tensile Stresses in Steel, 
Nee De i 


Compression stresses in steel do not need to be computed, because 
for satisfactory fc they are always lower than the allowable stresses. 

The values of Ca and k in the above formulas depend upon known 
dimensions of the section and known eccentricity. They may be 
taken from the diagrams on pp. 938 and 939. 

For positive bending moment, the eccentric thrust is applied 
above the axis and the maximum compression stresses are at the 
upper edge of the section. 

For negative bending moment, the eccentric thrust is applied 
below the axis and the maximum compression stresses are at the 
lower edge of the section. 

If the bending moment, M, is in inch-pounds and 6 and h in 
inches, the stresses are in pounds per square inch. 

To get stresses 1n pounds per square foot, the bending moment, 
M, must be in foot-pounds and b and h in feet. 

General Formula for k.—The value of the constant Ca in Formula 
(124) depends upon the known values of n, h and p and the unknown 
value of k. | 

The value of k may be determined from the requirement that the 
bending moment, M, as expressed in Formula (122), be equal to 


DIRECT STRESSES AND FLEXURE 183 


Ne where N is given in Formula (121). Multiplying the Equation 
(121) by e and equating it to Equation (122), we have 


k? + 2npk — "4 





f.bd = sone] "(2)" 4 #(8) 8 
2 k BETES d) seal ate 
After simplification, the equation changes, to 
lh 2 
3 eee = 
ke +3(5 oa) + Gnpk® = 3npaq + 6np(S) . (127) 


This third degree equation may be solved by substituting 


es -(§-53 
ea Feo 1), 


whereupon the final equation will assume the form 
2 + Ciz.t+ Co = 0, 


where C; and C2 are constants depending upon e, h, n and p. The 
solution of this equation by Cardan’s formula is 


z= y/— C2 + VGC)? + Ci) + 4/— 302 — VC)? + GCa)e. 


Finally 
e€ Lh 
ie op (* we 5a): 


The solution of the equation as outlined above requires con- 
siderable time and is not adaptable for practice. To simplify the 
work, the values of k are given in the diagram on p. 938 for different 














values of p and 1 but for a fixed value of a, namely, a = : d. The 


value of k may also be found by trial as described below. 
Curves for k.—If it is desired to prepare a diagram for different 
values of a, the following simplified method may be used. The 


Equation (127) is solved for bs, giving the following formula: 
2 
—ke+ 1. phn + 6up(5) 


= a ee ieee oak | Oe 
3k? + 6npk — 3np 7 


Que 


With this formula, instead of assuming values of 5 and determin- 


ing the corresponding k by the elaborate method, the work is reversed. 


184 THEORY OF REINFORCED CONCRETE 


For each value of p, the value of k is assumed successively and the 
corresponding value of 5 computed from the simpler Equation (128). 


The curves may then be plotted and used for determining the values 
of k. 

Curves for Ca.—The values of Ca may be taken from the diagram 
on p. 939. . The diagram is also based on a = 2d. : 

Use of Diagrams for k and C,.—For the solution of any problem 
both diagrams are necessary. The value of k is found from one 
diagram, and for this value and the known ratio of steel, p, the value 
of C, is found from the other diagram. This value of Ca, inserted 
in Formula (124), gives the required unit stresses in concrete. 

The following problem is very common in practice: 

Given the dimensions of the member, b, h, and d.and the area of 
steel, As; the ratio of elasticity, n; the thrust, N; and the bending 
moment, M. 

Required the compressive stresses in concrete, fe, and tensile 
stresses in steel, fs. . 

The solution of the problem is as follows: 


8s 


Find value of p = ou The percentage of steel = 100p. Find 


value of e from e = a Determine the ratio “ Locate on diagram 


N 


for k the line corresponding to the determined value The inter- 


section of this line with the curve marked with proper percentage of 
steel gives the value of k. Use, in diagram (23), the determined 
value of k and the known percentage of steel for determining the 
value of Cz, which with Formula (124) gives the stress in concrete. 

The stresses in steel may be obtained from Formula (126), 
p. 182. 


Determining k without Curves.—The value of k for any condi- 
tion not covered by the diagram, for instance for different ve may be 


found by trial from Formula (128).. This work is simpler than solv- 
ing the third degree equation. A value of k is assumed first, and 
then checked by substituting it ‘n formula. If the resulting value of 
7 is substantially equal to the actual value of 2 the assumed k is 
satisfactory. If not, another value should be assumed and the work 


DIRECT STRESSES AND FLEXURE 185 


repeated. Usually, two trials will give a close agreement. In 
assuming the first value of k, the diagram on p. 938 will be found 
useful, as in most cases the actual conditions will not be very different 
from those assumed in preparing the diagram. 

Steel in Tension Face Only.—The formulas given in preceding 
pages apply to members reinforced with steel in tension and com- 
pression placed symmetrically in respect to the axis. Often, members 
are reinforced for tension only, in which case the formulas given 
below may be used. The section is shown in Fig. 63, p. 185. As in 
the previous case, the sum of all stresses must equal the thrust. 
Since the area of tension reinforcement is pbd and there is no com- 
pression reinforcement, the sum of all stresses is 


N = 3f-bkd — f.pbd. 
The relation between f, and f, is the same as given on p. 180. 


ad — kd Lek 
fo = fo = fe. 


Substituting this value in the above equation, 


— 2np(1 — k) 


k2 
N =fbd _ 


(129) 





Since all the reinforcement is near one face, the center of gravity of 
the combined section will not coincide with the center of the concrete | 
section. 





S Total. 
Compression 


SS okd 





i} A A 
Grainy A508 
Neutral Axis 








Fic. 63.—Section with Steel in Tension Face Only. (See p. 185.) 


Referring to Fig. 63 and assuming that the effect of the reinforce- 
ment equals n times the effect of concrete, the area of the combined 
section is bh + (n — 1)pbd. The distance of the center of gravity 
from the top of the section is equal to the moment of the combined 


186 THEORY OF REINFORCED CONCRETE 


section about the top edge divided by the total area. After the 
equation has been simplified the final formula for the distance of the 
center of gravity of the section from the top edge is 


7 1+ 2(n- vo(§) 





de eS 
Z ayes 
CLE (5) 
d 2 
eee a 14+ 2(n—- r(;) 
or d~ 2d ee 


ig i Ls Deg) 


Values of a may be found from Table 37, p. 940, for different ratios 


of steel, p. The moment, M, about this center of gravity may be 
found by multiplying the total compression by (d. — 3kd) and the 
total tension by (d — d:). 


M = 1fbkd(d. — thd) + pbdf.(d — d.). 


Expressing f, in terms of fi, the formula changes to 


ur =pbr|h(4 — Be) + pnt (1 —@)|- «80 


This value must equal the thrust, N, from Formula (129) multiplied 
by e. Thus, 


ae 3 ‘ ane 
jig oR ae Bs joi?| (7 is 3 | ie ie “(1 a alt 








2k 2\d k; d 
from which 
KF — 5K) + 2pn(1 — H(1 3 
NS : (132 
Ae k? — 2np(1 — k) en oss ) 


This equation may be used for preparing a diagram for k for dif- 
ferent values of p. Also, the value of k may be found by trial, as 
explained on p. 184 in connection with members with steel in ten- 
sion and compression. 

The stresses may be obtained by solving Equation (131) for fe. 


M 
fem 


game Bik ial REY Ea aA) 
Pd Wad (he Beare? eg tS aie 
ve Be at) tm (1 7) | 





HOW TO COMBINE THRUST WITH BENDING MOMENT. 187 


bri ye, Ad we ih 1—k d. 

By substituting C, = (3 ~ h)+ pn (1 ar the formula 
changes to | 

Maximum Compression Stresses in Concrete, 

M , 
Maximum Tensile Stresses in Steel, 
1— ih 
aye eta Paige ale Raye bs 


The value of C, and k may be taken from diagrams, pp. 940 and 941. 


HOW TO COMBINE THRUST WITH BENDING MOMENT 


In most cases where the formulas for direct stress and flexure 
are applicable, the bulk of the thrust is produced by one kind of 
loading and the bending moment by another kind of loading. For 
instance, in arches, the dead load produces very small bending but 
comparatively large thrust. The live load, on the other hand, 
produces small thrust but large bending moment. The thrust and 
the bending moment are therefore produced by two different agencies 
independent of each other. When the live load producing the bulk 
of the bending moment increases, the thrust and the bending moment 
do not increase in the same ratio. 

The same condition is found in column design. The column load 
is produced by the load on all stories above the column, while the 
bending moment is produced mainly by the floors connected with 
the column. Here again, an increase in bending moment can take 
place, due to overloading of one panel, without corresponding increase 
in the thrust. Again, the thrust depends upon the loading of the 
floors above. It is a maximum when all floors are loaded and the 
thrust consists of the sum of all the live and dead loads above the 
floor. It is a minimum when the load consists of dead load only. 
It is possible that a condition producing maximum bending moment 
will take place simultaneously with a minimum thrust. 

From the above, it is obvious that, to get the worst stress condi- 
tion, it is not permissible to combine the maximum thrust with the 
maximum bending moment, since they do not always occur simul- 
taneously. On the contrary, it is necessary to find a type of loading 
which will give the required factor of safety. 


188 THEORY OF REINFORCED CONCRETE 


In every design problem, the object is to produce a member with 
a certain factor of safety, that is, a member in which, for a load equal 
to the design load multiplied by the factor of safety, the stresses are 
equal to or smaller than the elastic limit of the material. 

In simple members, such as beams subjected to simple flexure, 
the computed stresses due to dead load are of the same character as 
those due to live load, and their magnitude is in direct proportion 
to the intensity of the loading. Also, the computed stresses for 
twice the design load, for instance, are twice as large as for the 
design load. Therefore, instead of multiplying the loads by the 
factor of safety and then working with stresses at elastic limit, design- 
ers use the simple expedient of establishing working stresses by divid- 
ing the stresses at elastic limit by the factor of safety. The problem is 
then solved by computing the bending moments for design load 
and determining the dimensions so that the stresses do not exceed 
the working stresses. This method simplifies the work, but it 
creates the wrong impression that the stress conditions at the design 
load are the determining factor. This is particularly misleading in 
reinforced concrete, where the actual stress conditions at the design 
load are no indication of the actual stresses at the load equal to 
design load times the factor of safety, even in simple structures. 
This is still more the case where it is necessary to combine stresses 
due to different types of loadings. 

In structures subjected to combined avtion, this simplification 
is not permissible without first making an adjustment between 
the load producing the bending moment and the load producing 
the thrust. As explained above, it is impossible to get a common 
multiplier for the two types of loading. The examples will make the © 
matter clear. 

Suppose that, in an arch, a certain cross section, normal to the 
axis, is 12 in. wide and 18 in. deep and that the normal thrust due to 
the dead load is Na = 35 000 Ib. and acts at the center of the section. 
At the same section the thrust due to live load is N; = 15 000 lb. 
and the bending moment M; = 150 000 in.-lb. 

If the thrust and bending moments due to live and dead load were 
added, the resultant thrust would be N = 15000 + 35 000 =50 000 
lb. and the bending moment M = 160 000 + 0 = 160000 in.-lb. 
From Formulas (94) and (95), p. 170, the stresses would then be 


, = 150-0001 ©7160 000% 6 
“12 16 = aaa 








= 939 + 948 Ib. per 6q..1m, 


MEMBERS SUBJECTED TO DIRECT TENSION AND BENDING 189 


The maximum stress would figure f. = 480 lb. per sq. in. and 
minimum stresses f: = — 16 lb. per sq. in. 

Since compression is not excessive and there is no appreciable 
tension. in the section, the design would appear to be satisfactory. 
However, this is not the case, as there is no factor of safety. Sup- 
pose that a factor of safety of 2.5 is required, i.e., that the structure 
must stand 2.5 times the design live load before failure. Let us 
apply this to this example. Taking 2.5 times the live load, the 
thrust and the bending moment due to the live load given above 
must be multiplied by 2.5. Thus, NM; = 37 500 lb. and M; = 400 000 
in.-Ib. The thrust due to the dead load, on the other hand, can 
never be exceeded and remains 30 000 lb. 

Thus, the sum of thrusts for live and dead load is N = 37 500 + 
30 000 = 67 500 Ib., and the bending moment is M, = 400 000 + 
0 = 400 000 in.-lb. The stress is 


67 500 6 X 400 000 


te = 12 xX 18 "42 x 182 = 313 + 620 lb. per sq. In. 


The maximum stress is fe = 933 lb. per sq. in., and the minimum 
stress is f; = — 307 lb. per sq. in. 

For a live load equal to 2.5 times the design live load, therefore, 
instead of a tensile stress of 16 X 2.5 = 40 lb., we actually get the 
excessive tensile stress of 307 Ib. per sq. in. It is evident that, for 
this condition, the tensile stresses actually developed are consider- 
ably greater than the first method would have indicated. 

From the above illustration, it is evident that, to get a proper idea 
of the stress condition to be expected in case of overloading, it is 
necessary to investigate the stresses for dead load plus the live load 
multiplied by a factor of safety. For such a condition, instead of 
using working stresses as the requirement, the stresses must not 
exceed the elastic limit in materials. This is more fully explained 
in the “ Arch Chapter,” in Volume III. 

The method to be employed in designing columns subjected to 
bending is discussed on p. 458. 


MEMBERS SUBJECTED TO DIRECT TENSION AND BENDING 


Members are subjected to direct tension and bending when they 
receive simultaneously a central pull, 7, and a bending moment, M, 
or when the pull, 7’, acts eccentrically. These two conditions are 


190 THEORY OF REINFORCED CONCRETE 


interchangeable, as explained with relation to members subjected 
to positive thrust and bending, since they produce the same stresses. 
The central pull, 7’, and bending moment, 7, may be replaced by an 


eccentric pull, 7, acting at an eccentricity equal to 7 The reverse 


is also true. 

A positive bending moment, i.e., a bending moment producing 
tension at the bottom and compression in the top, may be replaced 
by a pull acting below the axis. A negative bending moment may 
be replaced by a pull above the axis. | 

Two conditions are considered below: first, the whole section is 
under tensile stresses; second, part of the section is under com- 
pression. 

For Notation see p. 167. 

Whole Section under Tension.—Assume that a rectangular 
section, with bars arranged as in Fig. 64, is subjected to a tensile 


force, 7’, and a bending moment, M. The eccentricity 1s ¢ = 7F;- 


fi: 
Assume that the eccentricity is such that the whole section is under 
———— 





eh 


<i) 


—————— > + 
‘eae 





Tage an ae 





Eccentric Pull applied Equivalent Central Pull and 
below Gravity Axis to Positive Bending Moment 


Fic. 64.—Section Subjected to Direct Tension and Bending. (See p. 190.) 


tension. Since conerete cannot resist any tension, all stresses must 
be resisted by the steel. For equilibrium, the sum of all the stresses 
must equal 7. Therefore, 


—T=—(A\ fi +a 


From the figure, it is found that the stress in the reinforcement 
nearest the force, N, equals | 


ate 
Pj Aad. 





Af,=T (136) 


MEMBERS SUBJECTED TO DIRECT TENSION AND BENDING 191 


The stress in the reinforcement away from N is 


nsf. af UAT ite 4 en SG re 
The two equations are sufficient to determine the stresses carried by 
the respective tensile members. Since in each equation two values, 
namely, A, and f,, are unknown, it is necessary to make some assump- 
tion. Either the area of steel is assumed and the stress determined, 
or the stresses are accepted and the areas computed. 

If A, = A’,, the required area of steel is obtained from the follow- 
ing equation, 
Tate 


oh la 





(138) 


in which f, is the maximum allowable tensile stress in steel. The 


stress in the other section is f’, = are It is smaller than the 
allowable stress. 

If it is desired to use the same unit <iguedt in both rows of bars, 
i.e., fs = f’s, the area next to T is found from Equation (136) and 
fhe area away from T from equation 

F Eta e a. 
A’, = orcas oer (139) 

The stresses in the set of bars next to the thrust are always 
tension. From Equation (137) it follows that in the other set of 
bars the stresses are tensile for e smaller than a. For e = a, ie., 
when force 7’ acts in the center of gravity of the one row of bars, the 
stresses in the upper row are zero. For e larger than a, compression 
stresses will be developed in the section and the formulas are not 
applicable. 

Part of Section under Compression.— When the eccentricity, e, 
is larger than a, part of the section is under compression. 

Referring to Fig. 65, p. 192, and making the same assumption 
as given in connection with simple flexure on p. 126, the relation 
between the stresses in steel and concrete is as follows: 


Unit Compression Stress in Upper Steel, 


fe = nf(1 - 5): catia Renal a. 20) 


192 THEORY OF REINFORCED CONCRETE 
Unit Tensile Stress in Lower Steel, 


f, = if | eS eee 


The stresses may be determined from the principle that for equi- 
librium the sum of the stresses acting on the section must equal the 


pull, 7. The stresses resisted by the concrete are ee The 


A 
stresses resisted by the compression steel are A fee anf ee ra): 





Axis 








Fic. 65.—Direct Tension and Bending Moment. Part of Section in Compression. 


(See p. 191.) 
The stresses resisted by the tensile steel are A.f; = pepcct 
Adding these, we have | 
Re bial eee gh | 

fp ihe + A’.nfe\ 1 id Awmf-——.- ~ (142) 

The sign before 7’ is minus because the force is a pull. 
Assume that A’, = As = 5d we have 
tba k? + 2npk — np 
ey ph a kat (143) 





2 k 


This equation 1s the same for positive thrust and moment, except 
that 7’ is negative instead of positive. 

The second requirement is that the bending moment of all 
external forces about the center of gravity must be equal to the 
bending moment of all internal forces about the same axis. Taking 


MEMBERS SUBJECTED TO DIRECT TENSION AND BENDING 198 


the moment of all stresses about the gravity axis and expressing f; 
and f’s in terms of fc, the equation of the moment of internal forces is 


as 2| rp /a\? Kh _ Ke 
M = pa] "P(S) tae el ee ee 1) 


’This equation is identical with Equation (122) for positive thrust and 
bending moment. For equal M, the stresses, however, will be dif- 
ferent because the value of & is smaller for pull and bending moment 
than for positive thrust and bending moment. 

The value in square brackets may be called Ca, and the equation 
changes to M = f-bd?C., from which 


Maximum Compression Stress, 


M 
fc = Oba" (145) 
Maximum Tensile Stress, 
fh ip oe pl percha a Ary 


Value of k.—In the above equation, the value of C, depends 
upon the value of k. This can be determined by equating the value of 
M from Equation (144) to the value of Te obtained by multiplying 
Equation (143) by — e. 


Solving the equation for —, we have 


a 
d 


2 
—k3 + 15k + 6np (5) 
= cj hahaliampie ba ican GE iy 


3k? — 6npk + 3p 





Qua 


_ This equation is similar to the Equation (128) for positive thrust 
and moment. The difference is in signs. 

The direct equation is an equation of the third degree. It is 
easier to find k by trial, as explained on p. 184. In assuming the first 
value of k, it must be remembered that it is much smaller than the 
k in ordinary beams. 

Use of Formulas.—In practice, the problem usually is to deter- 
mine the stresses in steel and concrete for known dimensions and 
given values of T and M. 


In solving the problem, the value of e is found first, from e = - 


194 THEORY OF REINFORCED CONCRETE 


Next, the value of k is found by trial from Formula (147) for the 
known value of - Finally, the value of k is substituted in the formula 


for C, and the values of fc and f, are computed. 
If dimensions b and d are in inches and M in inch-pounds, the 
stresses are in pounds per square inch. 


FLAT SLAB THEORY 


Flat slab construction is statically indeterminate, 1.€., the static 
equations of equilibrium are not sufficient to determine external 
shears and bending moments in the various parts of the slab. The 
problem can be solved only by consideration of the deflection of 
the slab and the relation between the deformations and the stresses. 
This makes the bending moments dependent not only upon the 
loads, as in statically determinate structures, but also upon the 
elastic properties of the slab and the relative stiffness of the different 
parts. The purely mathematical problem is made difficult because 
the material composing the slab is not homogeneous. It is also 
complicated by the fact that the slab deflects in two directions and 
that the equations for deflection are functions of three variables, 
x, y, 2, instead of two variables as is the case with continuous beams. 
The solution of the equations cannot be obtained by integration but 
by approximate summation. The actual solution is still further 
complicated by the uncertainty of the application to concrete con- 
struction of the formulas based on the assumption of homogeneous 
slab, and also by the uncertainty of the value of Poisson’s ratio. 
Furthermore, it has been observed that, in a slab under load, a read- 
justment of stresses takes place for which it 1s not possible to make 
any allowance in the analysis. 

Statical Limitations.—It is well known that, in continuous beams 
or slabs, for uniformly distributed load, the numerical sum of the 
positive bending moment in the center and the average of the nega- 
tive bending moments at the two supports is equal to the maximum 
positive bending moment (which is at the center) for the beam or 
slab if considered as simply supported. If — Ma and — M, are 
negative moments at the supports a and b, and M, is the positive 
moment at the center of the span, then | 


Mat My 4 yy, = jut 


FLAT SLAB THEORY 


195 


Applying the same rule to flat slab construction, Mr. Nichols 8 
demonstrated that in an interior flat slab panel, shown in Fig. 66, 
p. 195, the sum of the negative bending moment along section 1-1 
and the positive bending moment along section 2-2 equals static 


i 1-1= Section of Critical Negative 
7 ee Bending Moment 


Bending Moments act at right angles to Sections 





t\ 


ay 
Ne ae 
2 


l= Length of span in the direction of Bending Moments 


L= Length of span at right angles to | 


Note that the designation of | and depends upon the 
direction of Bending Moment and not upon relative 
length of spans. 


Fra. 66.—Moment Sections in Flat Slab Panel. 


(See p. 195.) 


bending moment in the center of a freely supported slab with a 
span equal to [(1—3c). This is expressed by the following equation: 


Let w = unit live and dead load; 
1 = side of a square panel; 
c = diameter of the column capital; 


M, = negative bending moment along the edge of the panel, 


section 1-1. 


M2 = positive bending moment along the center of panel, 


section 2-2. 


8 “Statical Limitations upon the Steel Requirement in Reinforced Concrete 


Flat Slab Floors,” by John R. Nichols, Trans., Am. Soc. C. E., Vol. 77. 


196 THEORY OF REINFORCED CONCRETE 


Then 
Sum of Bending Momenis, 


In0 Va 
My + Mz = uP (1 - 35) 1 hn SESE 

Distribution of Moments.—The above formula does not show 
how to distribute the bending moments between the positive and 
negative bending moment. The distribution can be determined 
only by means of equations taking into account the deformation 
of the slab. This problem was treated by Mr. Westergaard.? In 
his analysis, Mr. Westergaard used the results obtained by N. J. 
Nielsen for uniformly loaded square panels extending indefinitely 
in all directions and supported upon point supports. These results 
were obtained by dividing each panel into one hundred elementary 
squares and computing the deflection of each corner of each ele- 
mentary square. The moments then were determined from deflec- 
tions by the summation method. Since Nielsen’s results were 
not directly applicable to flat slabs supported along a column cap- 
ital, they were modified by Mr. Westgaard by the introduction at the 
column head of the so-called ring loads, consisting of a combination 
of an upward load uniformly distributed along the circumference 
of the column head, with an equally large downward load applied 
at the center of the column head. The two loads balance, but 
produce a bending moment at the column head. A ring load of 
the proper intensity at each point support, combined with a cer- 
tain uniformly distributed bending moment applied at the edge of 
the whole slab (which includes many panels), and combined with 
the original uniformly distributed load and the reactions at the point 
support, will make the slab deflect in such a way that at circles 
marking the column capitals it will become horizontal and tangential 
to the deflection curve. 

The sum of the positive and negative moments is the same as 
found by Mr. Nichols, since both theories start with the same gen- 
eral assumption. The distribution of the moments along the vari- 
ous design sections is given in the table. The design sections 
referred to in the table are those shown in Fig. 66, p. 195. 

Application of the Theoretical Moments.—The above bending 
moments and their distribution are only of theoretical significance. 


9 Moments and Stresses in Slabs, by H. M. Westergaard and W. A. Slater. 
Proceedings, American Concrete Institute, p. 415, Vol. XVII, 1921. 


FLAT SLAB THEORY 197 


In actual construction, the stresses in steel are much smaller than 
would be obtained from the above formulas. The reasons are: 

1. The Poisson’s ratio in the formulas was assumed as equal 
to zero. ‘This is not the case in an actual slab. 


Percentages of Sum of Positive and Negative Moments Resisted in Sections 
Shown in Fig. 66 


Computed by H. M. Westergaard. 


Results of analysis of a square interior panel of a uniformly loaded homo- 
geneous flat slab. 
The sum of positive and negative moments is approximately equal to 


i hbo eae 
es 31) 


c = diameter of column capital; 1 = span; W = total panel load; Poisson’s 
ratio = 0. 





J 
l 

















Description of Moments a __| Average 
| 
0.15/0.20,0.25'0.30 
( Across edge of capital. |31.8/37.140.2|42.7/20+80 ; 
| 
eG diside capital...... 16.5/11.3] 8.1] 5.7128—80 © 
; ; Strips. ke 
Negative 
ihe ae | eetital eee 48.3/48.4/48.3/48.4 48 
| Middle Strip...:......0c..0s.. 17.0|16.7|16.6/16.3} 17 
| Total Negative Moment...... 65 .3/65.1/64.9/64.7 65 
Positi Pigglumin Gui... 4:5. ..... a 20.9/20.9!20.8/20.7 21 
ehh | Made stinp 13.8|14.0/14.3/14.6, 14 
Moments... : Leen) © if! 
Total Positive Moment....... 34.7/34.9/35.1185.3 35 




















2. A considerable amount of tension is carried by the concrete. 
Since this happens in every test and since, because of this, flat slab 
construction is able to carry, without signs of distress, loads under 
which it should have failed according to the theoretical formulas, it 
is clear that advantage should be taken of these factors. 


198 THEORY OF REINFORCED CONCRETE 


8 Considerable arch action takes place, which increases the 
total compression, but reduces the total tension in the slab. 

Since the theoretical formulas are not in close accord with the 
practical results, they were modified for practice, and thus brought 
into closer agreement with the results from the tests. 

Action of Flat Slabs.—To get an understanding of the action of a 
flat slab, consider its deflection under load. Figure 67 shows, by 


Rectangular Section 








Fic. 67.—Contour Lines for Flat Slab. (See p. 198.) 


means of contour lines, the deflection of one panel taken from a con- 
tinuous floor, where it and the surrounding panels were uniformly 
loaded. The contour lines are curves connecting points of equal: 
deflection, i.e., points which, after loading, deflect an equal distance 
below the original level of the slab. The points: were obtained by 
drawing cross sections along the edge of the panel and along a diag- 
onal section, and plotting their deflection curves. This curve of 
deflection, just as for continuous beams, is determined by the require- 
ments that the tangents at the support and at the center of the 


FLAT SLAB THEORY 199 


span shall be horizontal and that the points of inflection shall be 
at about one-fifth of the net span. After the deflection curves have 
been drawn the contour lines were plotted from the deflection curves. 
From the contour lines it is evident that a flat slab, after deflection, 
assumes the shape of an umbrella at the column head, of a trough 
between the columns, and of a saucer in the central portion. 

Points of Inflection.—Since the bending moments change from 
negative at the column head to positive at the center of the slab, 
there must be a line of zero moment, or a line of points of inflection 
surrounding each column. ‘This line of points of inflection divides 
the slab into circumferential cantilevers concentric with the col- 
umns and firmly clamped to them, slabs extending between adjacent 
columns and supported on both ends, and square or rectangular 
panels supported on four edges. 

Stresses at Column Head.—It is evident from the contour lines 
and lines of deflection that the portion of the slab at the column 
head, which acts as a circumferential cantilever, is subjected to a 
negative bending moment causing tensile stresses at the top and 
compressive stresses at the bottom. Assuming the umbrella shape, 
the slab undergoes deformation in two directions, namely, in the 
direction of the radius of any circle drawn on the slab around the 
column, and also along the circumference of the circle, because 
both the radius and the circumference are increased by the deflec- 
tion. The particles, therefore, are subjected to two stresses, radial 
and circumferential, acting at right angles to each other. The rein- 
forcement at the column head, therefore, must be placed in two 
directions, preferably radial and circumferential. 

As explained in text books on mechanics, when a particle is 
subjected to forces acting at right angles to each other, its actual 
deformation in any direction is smaller than if the force in this 
direction acted separately, because the longitudinal deformation 
due to one force is decreased by the lateral deformation produced 
by the force acting at right angles. The ratio of decrease, called 
Poisson’s ratio,!° is taken into account in fixing the constants 
recommended in the Chapter VI on Design of Flat Slab Structures. 

10 Tf a particle is subjected to compression, it shortens in the direction of the 
force, but at the same time it extends laterally. The ratio between the longi- 
tudinal shortening and the lateral extension is called Poisson’s ratio. It varies 
for different materials. If the particle is subjected to stress in one direction only, 
the Poisson’s ratio has no significance. It is important only when the particle 
is subjected to stresses at right angles to each other. 


200 THEORY OF REINFORCED CONCRETE 


Stresses in Central Portion—The central portion of the slab 
between the points of inflection 1s subjected to positive bending 
moment, causing tension below and compression above. ‘The por- 
tion of the slab between adjacent columns develops stresses in one 
direction only, since the contour lines are practically perpendicular 
to the center line through the columns. The portion in the middle 
of the panel, on the other hand, is stressed in two directions, since 
the conditions there are somewhat similar to those at the column 
head, except that they are reversed. 

In the first case, the steel must be placed in one direction only, 
while in the other case, it must be circular or placed in two direc- 
tions. 

The bending moments recommended for design are given in 
Chapter VI. 


CHAPTER V 
REINFORCED CONCRETE DESIGN 


In this chapter are given the definite principles and rules used in 
the design of reinforced concrete structures. The matter is based 
on the two preceding chapters, one of these goes much more fully 
than the present chapter into the fundamental theory of reinforced 
concrete, giving formulas and their derivations for rectangular 
beams, T-beams, beams with steel in top and bottom, columns, and 
members under direct compression and flexure; while the other 
chapter describes the tests which verify both theory and rules for 
design. The working formulas which are necessary in actual design 
are taken up in the present chapter. Before using these final 
formulas, the designer should become acquainted with the deriva- 
tions already given so as to have a thorough understanding of the 
subject. 


RATIO OF MODULI OF ELASTICITY 


As given in Volume II of this treatise, the value of the modulus 
of elasticity of concrete depends upon the quality of the aggregates 
- used, the consistency, and the age. It varies also for different 
intensities of the stresses, but may be considered constant within 
working limits. The modulus of elasticity of steel being practically 
constant (see p. 203), the ratio of the modulus of steel to that of con- 
crete changes in direct proportion to the change in the modulus of 
concrete. ‘This ratio is designated by the letter n. 

In computations, it is advisable to vary the ratio according to 
the ultimate strength of the concrete. The ratios, n, recommended ! 
for use are: 


(a) For concrete having a crushing strength of 2 200 lb. per 
sq. in. or less, a value of 15. 


1 These values agree with the recommendations of the Joint Committee, 1924. 
201 


202 REINFORCED CONCRETE DESIGN 


(b) For concrete having a crushing strength between 2 200 
and 2 900 Ib. per sq. in., a value of 12. 
(c) For concrete having a crushing strength exceeding 2 900 
Ib. per sq. in., a value of 10. 


The value of 15 has been adopted in the British, German, and 
Austrian rules up to 1916. ‘The French rules for 1907 authorize a 
range from 8 to 15 according to conditions. For determining 
deflection of beams when using formulas which do not take into 
account the tensile strength developed in the concrete, a ratio of 
8 may be used. 

The effect of the ratio of moduli on the stresses in beams may 
be seen from the formulas and also from the tables. For a given 
beam with a definite amount of steel, the use of a higher ratio of 
moduli lowers the position of the theoretical neutral axis, and for a 
given bending moment decreases the stresses in concrete and increases 
the stresses in steel, the latter, however, in much smaller propor- 
tion. For the same unit stresses and bending moments, but dif- 
ferent ratios of moduli, the beam designed for the larger ratio will 
have a smaller depth, but at the same time a larger amount of steel. 
Therefore, in beam design, if a concrete richer than ordinary is used, 
the question of economy must be carefully considered. 

Modulus of Elasticity in Tension.—But few tests of modulus of 
elasticity of concrete in tension have been made, but these indicate 
that the value is probably the same as the modulus in compression. 

Significance of Yield Point, of Steel in Reinforced Concrete.— 
In reinforced concrete construction, the yield point of steel marks 
the failure of the structure. The ultimate tensile strength of steel 
is important only as far as it indicates the quality of the material. 
It can never be reached in reinforced concrete structures, as all 
concrete construction fails by compression soon after the yield 
point of steel is exceeded. Tests show that when the yield point 
of steel in the beam reinforcement is reached, the beam sags, and the 
cracks widen and extend upward into the compression zone. Any 
small increase in loading beyond the load producing yield point 
increases the width of the cracks considerably and extends them up 
into the compression zone. When the load is still further increased, 
the compression zone becomes finally reduced to such an extent 
that the beam fails by crushing of the concrete. This type of failure 
takes place even if the beam is very strong in compression. As the 


FORMULAS FOR DESIGN OF RECTANGULAR BEAMS 203 


failure always follows closely the yield point in steel, the strength 
of the beam is based upon the yield point of the reinforcement, and 
not upon its ultimate strength. 

Modulus of Elasticity of Steel—The modulus of elasticity of 
all grades of steel is substantially the same, namely, 30 000 000 lb. 
per sq.in. It does not vary with the strength of steel, as is the case 
with concrete. This is unfortunate, as steel with high modulus 
of elasticity would be particularly serviceable for reinforced con- 
crete, because the higher the elastic limit, the smaller is the deforma- 
tion under any given loading. 


FORMULAS FOR DESIGN OF RECTANGULAR BEAMS 


Reinforced concrete beams must have breadth and depth suffi- 
cient to keep the compression stresses within working limits, and 
enough tensile reinforcement to take all the pull without exceeding 
the working stress of the steel. Rules for this are given in the simple 
formulas which follow. The bond stresses must be investigated 
(see p. 260), and also inclined or vertical reinforcement may be 
required, as stated in connection with diagonal tension, pp. 241 to 
260. Continuous beams also require tensile reinforcement over 
the supports, as described on pp. 285 to 290. 

For derivation of formulas, see chapter on “ Theory of Rein- 
forced Concrete.” 

Balanced Design of Rectangular Beam.—The design of a rec- 
tangular beam is balanced if its concrete dimensions are determined 
by Formula (1), (1a) or (2); and the amount of steel by Formula (3), 
using constants corresponding to the specified unit stresses. In 
such beams unit stresses in both materials reach their maximum 
working value simultaneously. 

Unbalanced Design of Rectangular Beam.—The design is not 
balanced, if the depth is larger than required by Formula (1), or the 
amount of steel larger than required by Formula (3). See also p. 206. 


Let d = effective depth of beam; 
6 = breadth of rectangular beam; 
A, = area of steel in tension; 


OLA 
p = steel ratio, bd 
M = bending moment or moment of resistance; 


204 REINFORCED CONCRETE DESIGN 


C = constant in table, pp. 205, 880 and 881. 
R = constant in table, pp. 205, 880 and 881. 


Formula’ for Design of Rectangular Beam.—The maximum bend- 
ing moment, M, due to the loads, having been computed, the breadth 
of the beam, }, is assumed; then the depth of the beam, d, and the 
amount of steel, A , are found from the following formulas: 







Td f, bed 


Total Compression 








=) ' ‘ 1 
| : 
| ! 
Neutral is | 
, 7 bo 3 | 
Axis Sie | 2 
2 ! Hl 
,& ' ' 
is 
| ~ Total Tension H : 
= wickeresea Y_.- 
f ‘ podfs orn 
oot es ee ere ee baa 
n 


Fic. 68.—Resisting Forces in a Reinforced Concrete Beam. (See p. 204.) 


Depth of Beam for given 6, 


pean 5). (ld caeelotes cae ne santas 


Width of Beam for given d, 





M 
b = Rd2’ e e e e e e e e e e e e e (2) 
Area of Steel, 
M 
Avs pod, “se A or A, = safe (3a) 


In computing the area of steel, Formula (3) should be used only 
with the values of d found from Formula (1) or (1a). If larger d 
is used, Formula (3a) should be used for finding steel area. 

For ordinary cases, the value of 7 may be taken as . 

If M is in inch-pounds and 6 in inches, d is in inches and As in 
square inches. The constants C, p, and R may be taken from the 
table on p. 205, those values being selected which correspond to 


FORMULAS FOR DESIGN OF RECTANGULAR BEAMS 205 


the working stresses in steel and conerete and to their ratio of elas- 
ticity. 

The following table gives the values of constants for selected 
stresses: 


Cons‘ants in Beam and Slab Design * 


nde on 
M M 
For use in beam formulas d = oxi and d =e and in slab formula d = 


C;V M . (See pp. 204 and 208.) For additional values, see table on p. 880. 












































| Allowable | 
Unit Stresses, 
Lh. per Sq. In. 
Propor- |Ultimate| k 7 D C Ce TG 
tion /|Strength | 
Steel | eos 
f crete 
oy 
Pete A eB OOO SLO Fe: (1200) 0.429; 0.857/0.0161) 0.067|0.019) 221 
1:13 3 2 500 | Hes hehe | 1000} 0.429) 0.857/0.0134| 0.074'6.021|} 184 
1:2 :4 2000 /15 }| 16000 800 | 0.429! 0.857/0.0107| 0.083/0.024) 147 
Peeea Die OOO TB ile cect 640| 0.375) 0.875)0.0075) 0.098'0.028) 105 
NGS Sor Bek. er re 520} 0.328) 0.891/0.0053) 0.115'0.033) 76 
feerme 2 a OO. i101 ii” 1 200 | 0.400) 0.867/0.0133; 0.070.0.020; 208 
Peviekias ver) 2b F 000 | 0.400! 0.867;0.0111) 0.076/0.022| 173 
1:2: 4) 2000 |15 7| 18000 800} 0.400; 0.867/0.0089 0.085)0 024; 139 
Poe GO LOO Bl... . 640} 0.348) 0.885)0.0062| 0.10110.029; 99 
ee Ol Le Lor) | eicue. . 520} 0.302} 0.899'0.0044| 0.118/0.034| 70 








| 


Important Note: The values in the above table can be used only when 
special care is taken to produce concrete of strengths assumed in the table and 
when the strength is checked by field tests. If no field tests are made, use in 
design the constants for the next lower mixture. Thus for 1 : 2 : 4 concrete use 
constant corresponding to 1 : 23 : 5 concrete. 


Example 1.—What depth of beam and what area of steel are required for a 
freely supported beam having a span of 18 ft. and supporting a load of 600 lb. 
per running foot. 2000 lb. concrete and structural grade steel are used. The 
stresses are fe = 800, fs = 16 000, n = 15. 





1 

* = 

: 18 Ta hk a A tl i RR i licen ep 
nf A Ast (Redhat GUMNAR TS} ie 


206 REINFORCED CONCRETE DESIGN 


wl? 600X 18 X18 X12 
Solution.—Bending moment, M = = x 12 = ; 

in.Ib. From table above, for specified working stresses, the constants are C = 

0.083 and p = 0.0107. Assuming a breadth b = 8 in. and using Formula (1), 


291 600 : 
d = 0.083 3 =15.8 10. 


With 2 in. of concrete below the center of steel, the total depth of beam is thus 
17.8 in. In practice, a depth in even inches should be adopted. 

The area of steel, from Formula (3), is A = 0.0107 X 8X 15.6 = 1.348q. in., 
thus requiring three j-inch round bars, or their equivalent. 





= 291 600 





Value of fs, f-, and p are Interdependent.—The stresses, fs and fe, 
and the ratio, p, are interdependent. To any pair of values of fs 
and fe, there corresponds only one value of p. With f; and f, given, 
the corresponding ratio, p, must never be exceeded, else, if the 
stress in the steel f;, were maintained, the stress in the concrete 
would be increased beyond the allowable working stress, fc. 

If, for resisting the bending moment, it is necessary to use a 
larger ratio of steel than the value p corresponding to the adopted 
working stresses, and at the same time maintain the stress in steel, 
f;, and stress in concrete, fey the excess tensile steel must be balanced 
by compression steel, as discussed on p. 232, or else concrete of 
richer mix should be used. | 

Sometimes the ratio, p, is increased to reduce bond stresses. In 
such a case, however, the stress in steel is reduced correspondingly, 
so that the compression stresses are not affected. 


Example 2.—To the stresses fs = 16 000 and f, = 650, specified by the New 
York Code, corresponds a ratio of steel p = 0.0077. If a ratio of p= 0.01 
were used and the steel were stressed to 16 000 lb. per sq. in., the corresponding 
stress in concrete would be 770 lb., instead of 650 Ib., per sq. in. To maintain, 
with a ratio p = 0.01, the stress in concrete fe = 650 lb. per sq. in., without 
adding compression steel, it would be necessary to limit the stress in steel to 
13 000 Ib. per sq. in. This shows that, with too large a ratio of steel and no 
provision for additional compression, the full working value of steel cannot be 
utilized and therefore the beam is not economical. 


Formulas to Review Rectangular Beams Already Designed.— 
The following formulas, the derivation of which is given on p. 180, 
may be used to review a beam already designed. 

f. = compressive unit stress in concrete, in pounds per 
square inch; 
f, = tensile unit stress In steel, in pounds per square inch; 


FORMULAS FOR DESIGN OF RECTANGULAR BEAMS 207 


= breadth of beam in inches; 
d = effective depth of beam in inches; 
k = ratio of depth of neutral axis to depth of beam, d : 
j = ratio of distance between the centers of compression 
and tension to depth of beam, d; 
. k , 
1a, = a(t — 5) = moment arm, or distance between the 
centers of compression and tension; 


A, = area of cross section of steel in square inches; 
p = steel ratio = A, + bd; 
M = bending moment in inch-pounds; 
n = ratio of modulus of elasticity of steel to that of concrete. 
Then | 
: A; rte (on \e : k 
p=5q @) k=Vinp FOP — np () j=1-% ©) 
Stresses in Steel and Concrete, 
M 2M 
emery: fe bbjke POP CY 


The value of p is figured first, then k and j are computed or taken 
from the table on p. 205, and substituted in Equations (7) and (8). 

Neither the allowable tension in steel nor the allowable com- 
pression in concrete should be exceeded. For explanation see 
p. 206. 

Approximate Formulas.—For rectangular beams designed with 
the stresses ordinarily used, the moment arm, jd, is about 7d and the 
above formulas may be expressed as 


Stresses in Steel and Concrete, 
M _ 6M 


ee!) kaa (10) 


Tables and Examples. 

Tables for determining the Rinintatsenke and loading of rectan- 
gular beams are given on pp. 884 and 885, and the methods of prac- 
tical computation and details of design are illustrated in Example 1, 
p. 578. T-beams are treated on p. 215. 

The selection of bending moments for use in design of continuous 
beams is treated on p. 275. 

Factor of Safety.—Full explanation of the Pector of safety is 
given on p, 125, 


208 REINFORCED CONCRETE DESIGN 


DESIGN OF SLABS 


A slab, so far as computation is concerned, is a rectangular 
beam. The dimensions and stresses, therefore, can be obtained 
by the formulas given for rectangular beams. 

The bending moment is usually figured for a 1-ft. width of 
slab. The formula for depth of slab can be simplified by com- 
bining the selected value of 6 = 12 in. with the constants given for 


rectangular beams, changing formulas as given below. In the 


formula for required area of steel, As, it is most convenient to assume 

a width of slab, 6 = 1 in. The formula then gives the area of steel 

per inch of width of the slab, and the spacing of the bars can be 

readily determined by dividing the cross-sectional area of a bar by 

the determined area per inch of width of slab. 
Using notation on p. 203, and making 


M = bending moment in inch-pounds per ioot of width of slab; 
A, = area of steel in square inches per inch of width; 
C, = constant based on these units; 

the Formulas (1) and (3), change to 


Depth of Slab, 


I 


| — 0.290/M = Civ M (inches)) So yaaa) 
Area of Steel, 
A, = pd (per inch of width), . .... - (12) 


The table on p. 205 gives the values of constants, C1 and , for 
conerete of selected proportions. If larger depth of slab is used 
than required by this formula, the area of steel may be found from 


A, (per inch of width), . . . . (18) 


wy 
 «2jdf. 
The use of these formulas is illustrated in Example 1, p. 579. 

Tables and Examples.—Tables for determining dimensions and 
loading of slabs can be found on pp. 888 to 893. Table 6, on p. 886, 
gives dimensions and reinforcement for slabs for different live loads 
based on stress in concrete, f. = 800, stress in steel, f, = 16 000 
and 18 000 respectively, and ratio of elasticity, n = 15. Examples 
and details of design are given on pp. 210 to 579. 

Continuous Slabs.—In slabs continuous over two or more sup- 
ports, provision should be made for negative and positive bending 


DESIGN OF SLABS 209 


moments. For substantially equal spans and uniform loading, the 
following bending moments should be used: 


Maximum positive bending moment near the center and negative 


bending moment at the support should be M = ne for interior 


spans and M = ue for exterior spans. The negative bending 


2 
moment at the end support should be not less than M = vs The 
span length in the formulas is the span between the edges of the 
supports. If in the above formulas w is in pounds per square foot 
and / is in feet, the moments are in foot-pounds. To change the 
moments to inch-pounds, multiply by 12. 

Reinforcement at Supports.—To provide proper negative bending 
moment reinforcement at the supports, some of the bars (from one- 
half to five-eighths of the total area of bars) may be bent up and 
extended over the support into the next span. There they may 
terminate at the top about 6 in. beyond the point of inflection in 
this next span, or, if of 2-span length they may be bent down there 
and serve as positive bending moment reinforcement in the next 
span. Since the required area of steel at the support is equal to 
the required area in the center, and not all of the bars are bent up, 
the balance of the required area must be made up either by bent-up 
bars extended from the adjoining span or by extra short bars. Sev- 
eral arrangements of slab reinforcement are shown in Fig. 69, p. 210. 

Points of Bending Slab Steel—The point of bending should be 
selected so as to satisfy best the positive and negative bending 
moments. The authors recommend bending up the bars at a dis- 
tance from the support equal to one-fifth of the net span, J, plus one- 
half the depth of the slab. For thick slabs, the point may be moved 
somewhat farther from the support. 

The bend in the bars should be made around a pin of a diam- 
eter at least three times the diameter of the bar. Sharp bends may 
be injurious to the concrete. The bent portion should make, with 
the horizontal, an angle of from 30° to 45°, depending upon the 
ratio of the thickness of the slab to the span. The lower value 
should be used for thin slabs and the upper value for thick slabs. 
In exceptional cases, for very thick slabs, the angle of the bend may 
be made larger. 


210 REINFORCED CONCRETE DESIGN 






f] 
'- | Bent 









ra Min. 0.11, , Straight ~+ 
= Pie ae 
les wi | | | 
[7 | pies a « 
5 anetones = 
1 1 ie ' - = 
i. 2 Sig S Max> | 8 Bending Sketches 3 al 
Q wm . ' Q, 2 @ 
} 2 ASwwel \ BS a ES 
j ara he SS re We--- J ---- Sisco nae ne Ebr a 
| ots NS | ©), ee oS po 
NIE Bia ee Nis Nig \ 
Bs l 3! ioe &| l 
| | Aig i ! \ Bent | | 1 
accent ears, i] 


i} 
| <1--Beam 


“Straight <A win, 0,1 t-.? 


Alternate-<<_- 








‘<{-Beam 


}cheme 2° Combination of One Straight and Two Bent Bars with Four Double Benas. 


_--- Beam ---- Beam Beam--.. 





Straight Top ---Beam 










Straight+-* ie S.-- Straight Top---’ 
Min 0.151 Ra Plan 


Scheme 3 Combination of One Straight, Two Bent Bars, and Straight Bars on Top. 


eepmard ers ye 
0.214 6k >|  K-Sf 0.214 6" 





Scheme 4: Long Straight Bars on Bottom Short Straight Bars’on Top. 


Fig. 69.—Arrangement of Slab Reinforcement in Continuous Slabs. 
(See p. 209.) 


DESIGN OF SLABS PAM 


Cross Reinforcement of Slabs.—Cross reinforcement, that is, 
bars at right angles to the principal reinforcement, is customarily 
used to prevent shrinkage and temperature cracks. The amount of 
steel to use for this is usually selected somewhat arbitrarily, a cross- 
sectional area of bars equivalent to 0.2 per cent to 0.3 per cent 
(p = 0.002 to 0.003) of the cross section of the slab being usual in 
practice. 

Reinforcement over Girders.—If a girder is parallel to the prin- 
cipal reinforcement of the slab, short bars should be placed at the 
girder transversely, near the top of the slab, not only to stiffen the 
T-beam (see p. 38), but also to provide for the secondary negative 
bending moment produced by the bending of the slab next to the 
girder. ‘This reinforcement is necessary even when the beam is 
simply a small stiffener. (See p. 223.) 

The authors recommend that the amount of such reinforcement 
should not be less than 0.3 per cent of the effective area of the slab. 
The bars should be placed near the top of the slab, not farther 
apart than 18 in. Their length should not be less than two-thirds 
of the total effective width of the flange as given on p. 218. This 
recommendation agrees substantially with the 1924 Joint Committee 
recommendation. : 

Bar Spacers.—To keep the bars in place and a proper distance 
above the forms, the use of proper spacers is recommended. ‘There 
are numerous types of spacers on the market, which serve not only 
to keep the bars in place, but also to keep them a proper distance 
above the forms. If no spacers are used, the bars should be securely 
tied to the transverse reinforcement. 

Computing Ratio of Steel in Slab.—The ratio of steel in a slab 
may be found by dividing the cross section of one bar by the area 
between two bars, this area being the spacing of the bars times the 
depth of steel below top of slab. For example, a slab with steel 4 in. 
below the top and 4-in. round bars spaced 6 in. apart has a ratio, 


p = ie = 0.0082, or 0.82 per cent steel. 

Square and Oblong Slabs Supported by Four Beams.—When a 
slab is supported by four beams and its length does not exceed 13 
times its width, the loads will be carried by the slab to all four 
beams, and therefore the slab must be reinforced in two directions, 
as shown below. 


212 REINFORCED CONCRETE DESIGN 


The load carried in either direction can be determined from the 
following formulas: 


Let w = total unit load; 
we = unit load carried by the short span; 


wz, = unit load carried by the long span; 
L = length of slab; 
B = width of slab; 
Then 
apie (5 si 5) : (14) 
(ey ? , 
and 
Wr = (1.5 — 5) . (15) 
B 


The following table gives the ratio of the unit load, we, carried by 


the short span for different ratios of ak 


Load Carried by the Shorter Span. (See p. 212.) 


























Ratio of Length | qoad Carried by Ratio of Length | 149d Carried by 
to Breadth of : to Breadth of 
L the Shorter Vk the Shorter 
Slab B Span, we Slab B Span, we 
Jet Se ert | 4 ee Ee ee 
1.00 0.50w 1.30 0.80w 
1.05 0.55w a oe te 0.85w 
1AM, 0.60w 1.40 0.90w 
VO: 0.65w 1.45 0.95w 
1.20 0.70w 1.50 1.00w 
de 0. 75w 





It must be noticed that the shorter span carries the larger pro- 
portion of the load. 

After the proportion of the load is determined, the bending 
moments are found as for slabs reinforced in one direction, and the 
dimensions or stresses are found by the ordinary formulas. The 
thickness of the slab, of course, is governed by the larger bending 
moment of the two. 

The above proportion is based on the assumption that the con- 
ditions of restraint on all four sides are the same. It does not apply 
if the slab is continuous in one direction and simply supported in 
the other direction; in this case the slab in the continuous direc- 


DESIGN OF SLABS 213 


tion, will resist a much larger proportion of the load, because it takes 
a larger load to deflect a continuous slab an spueTemi ahs equal to the 
deflection of a simply supported slab. 

Most Economical Design of Rectangular Beam.—To get the most 
economical design, the ratio of width of beam to depth should be as 
large as practicable. For any given width of beam best economy is 
obtained by the use of the balanced design of beam (see p. 203). 
No economy results from using a larger depth than required by’ 
Formula (1), p. 204, because the cost of additional concrete is larger 
than the saving in the cost of steel. 

Economy of Rich Mix of Concrete in Slabs.—Under ordinary 
conditions no economy results from the use in slabs of concrete of 
richer mix than the leanest permissible in reinforced concrete, for 
the reasons given below. (See p. 4 for permissible mix.) 

The use of richer mix of concrete results in smaller depth of slab 
with consequent reduction in the amount of concrete, but it requires 
a larger amount of reinforcement than necessary for the leaner mix. 
While the amount of concrete in cost for rich mix is smaller than for 
lean concrete, the difference in cost due to higher unit cost of rich 
_ mix, is small. The saving in cost of concrete is more than offset by 
_ the extra cost of steel. 

As an illustration, compare the cost of slab required by 
M = 40000 in.-lb. using 1: 2:4 and 1: 14:3 concretes with the 
unit stresses given in table on page 205. Following unit prices ? 
should be used: 1: 14 : 3 concrete—49 cents per cubic foot; 1: 2:4 
concrete—45 cents per cubic foot; steel—5 cents a pound. 

The concrete dimensions and the required area of steel are found 
from Formulas (11) and (3). 





Mix Effective Depth Total Depth Area of Steel 
Gh es d=0.021V 40 000=4.2 in. h=4.2+1=5.2in. A,=0.0133 X4.2 X12 
= 0.672 sq. in. 
fee 2 4 d =0.024-V 40 000=4.8in. h=4.8+1=5.8in. A,=0.0107 X4.8 X12 
! =(0.615 sq. in. 
Volume and cost of concrete: 
Mix Volume X Unit Cost =Cost per sq. ft. of floor 
B.2 X 12 | 
Pig 3 ——_— = 0.48 cu. ft. X 49= 21 cents 
144 
2 
Ree 34 cee = 0.48 cu. ft. X 45=21.6 cents 


Saving due to rich mix = 0.6 cents per sq. ft. 


2 Prices prevailing in year 1925. 


214 REINFORCED CONCRETE DESIGN 


Weight and cost of steel: 
Mix Area of Steel Weight of Steel X Unit Cost = Cost of steel per sq. ft. 


of floor 
1: 12:3 0.672 sq. in. 2. 3ilbr Xx So = Ts eo 
1s 2.3 46280,.615. seein: 2.1 lb. XS 10centa 


Extra cost of steel = 1 cent per sq. ft. 


Since there is a saving in concrete due to rich mix of 0.6 cent 
and an extra cost of steel of 1.0 cent, the difference is 0.4 cent per 
sq. ft. in favor of leaner mix. Hence rich mix not economical. 

Economy of Rich Mix of Concrete in Rectangular Beams.— 
Under some conditions the use of rich mix in beams is economical. 
It is particularly advantageous for very long spans or where heavy 
loads are to be carried with limited headroom. Often, the reduction 
in dead load resulting from the use of rich concrete is of advantage. 

Rich mix of concrete permits reduction of the concrete dimensions 
of a beam. Either the depth may be reduced, keeping the width 
same as for leaner mix, or the depth may be used same as for leaner 
mix and the width reduced. 

In the first case the conditions are identical with those explained 
in connection with slabs and no economy would result. In the 
second case distinct economy may result from the use of rich mixture 
as evident from example below. 

As an illustration compare the cost of beams designed for 1: 15: 3 
and 1: 2:4 mixes for a bending moment M = 720 000 in.-Ib. when 
the depth of beam is limited to h = 22 in. using stresses and constants 
in the table on p. 205. | 

Since the total depth is h = 22 in., the effective depth d = 22 — 
3.5 = 18.5 in. This allows for 2 in. fire proofing and two layers of 
steel. Since the depth is fixed, the width of beam will be obtained 
from Formula (38). 


Mix Width of Beam Volume of Concrete per lin. ft. 
720 000 11.4 X 22 
1:14: 36 = ——— = 11.4in. ——— = 1.74 cu. ft. 
: 184 X 18.52 144 et 
720 000 14.3 X 22 
1:2:4 b= —— = 14.3in. ———— = 2.18 cu. ft. 
147 X 18.5" rE 144 ee 


Comparison of cost: 
Mix Volume of Concrete X Unit Cost = Cost per lin. ft. 


Lely 1.74 cu. ft. X 49 cents = 85.3 cents 
122%4 2.18 cu. ft. X 45 cents = 98 0 cents 


DESIGN OF T-BEAM 215 


The saving in favor of rich mix is 98.0 — 85.5 = 12.5 cents per 
lineal foot. The amount of steel is the same in both cases. An 
additional saving will be due to the reduction in dead load which 
amounts to (2.18 — 1.74) x 150 = 66 lb. per lin. ft. 


DESIGN OF T-BEAM 


The formulas given below are sufficient to design or review a 
T-beam for any given span and loading. The derivation of the 
- formulas is given on p. 134. Formulas for bond stresses and for 
diagonal tension are given in subsequent pages, and the design at 
supports for negative bending moment is also treated. | 

How T-Beams are Obtained.—If a beam is built at one pouring 
with the slab, and precautions are taken to prevent separation of 
the slab and beam, the portion of the slab near the beam may be 
considered as part of the beam. The beam then assumes the shape 
of the letter T and is called a T-beam. 

Figure 70 shows a cross section through a floor. The section 
of the slab considered as part of the T-beam is indicated. 











: atgle 


Fig. 70.—Section Showing T-Beam in Floor. (See p. 215.) 


- NOTATION 


Let 6 = breadth of flange of T-beam in inches; 
b’ = breadth of web of T-beam in inches; 
d = effective depth of T-beam in inches; 
jd = moment arm in inches; 
t = thickness of flange in inches; 
r = ratio of cost of steel to that of concrete; 
f, = compressive unit stress in concrete, pounds per square 
inch; 
f,; = tensile unit stress in steel, pounds per square inch; 
n = ratio of modulus of elasticity of steel to that of concrete; 
A, = area of tensile steel in square inches; 


216 REINFORCED CONCRETE DESIGN 


Dm = ratio of steel, a in T-beam for which stresses in steel 


and concrete simultaneously reach their maximum 
working value; 
constant for figuring minimum depth. 


Ca 


Formulas used in Design of T-beams.— 
Area of Web required by Diagonal Tension, 


I 


v(d — 5) equal or larger than 7 LEP) 


2 
Minimum Depth Required by Compression, 
M 
(Luz Ct (17) 
where C, is constant on p. 894 for fi, fs and -: 
For explanation see p. 219. 
Economical Depth, 
rM 6k 
where r is cost ratio of steel and concrete. 
Area of Steel, 
M 
As = a Ae As 19 
jaf a) 


where j is as given on pp. 221 and 897. 

This area of steel must not exceed the value found from the 
formula below, else compression stresses will be exceeded. 

Maximum Area of Steel Permitted by Working Compression 
Stress, fc, . 

? max. A, = Pmbd,.<-) G. yeiteleeee enn eneE 


on 


where pm is constant on p. 895 for f., fs, m and 7 

For explanation see p. 136. 

Area of steel used in design must not be larger than the above 
value, else compression stresses will be exceeded. 

Maximum Moment of Resistance of Beam Based on Working 
Stress in Concrete, fe, | 

. M = dCabt,~ 220 Be ee 

t 


where C, is constant on p. 894 for f., fs and 7 


For explanation see p. 186. 


DESIGN OF T-BEAM 217 


The maximum area of steel given in the formula just above 
corresponds to this maximum moment of resistance. 

The uses of the above formulas are explained below under proper 
headings. 

To Design a T-beam.— Design the slab. 

Determine bending moments and end shears in beam. 

Determine width of flange (see p. 217). If headroom is not 
limited, determine most economical depth (see p. 220). (In design- 
ing a number of similar beams, the eccnomical depth needs to be 
figured only for one beam, and estimated for the remainder.) 

Before selecting final depth of beam and breadth of stem, b’, see 
that the compression in concrete in the center and at the support 
and the shearing unit stresses do not exceed the allowable working 
stress (pp. 244 and 879). For continuous T-beam, the depth required 
at the support, where the T-beam becomes a beam with steel in top 
and bottom, may be the governing depth. 













r | dl : uy 2 Bhd... 


Total Compression 


Moment Arm : 


Total Tension 





nS 


Fig. 71.—Section of T-Beam. (See p. 216.) 


Figure amount of tension steel by Formula (19), p. 216. 

Design web reinforcement. 

For large T-beams, a saving in concrete may be effected by using 
the more exact formulas, which take into account the compression 
in stem below the flange. In such cases, preliminary A, for k and 
z may be found from Formula (19), and final A; from Formulas (19) 
to (24), p. 134. The problem is more easily solved by the authors’ 
simplified method given on p. 224, which gives exact results and 
requires much less work. 

Width of Flange.—The width of the slab, b, to use for the flange 
of the T-beam is selected by judgment on the basis of results from 
tests. In no case, of course, can it be greater than the distance 
between beams, The Joint Committee has recommended the 


218 REINFORCED CONCRETE DESIGN 


following rules, which are approved by the authors, for the effective 
width of flange: 


(a) The effective width shall not exceed one-fourth of the 
span length of the beam; 

(b) Its overhanging width, on either side of the web, shall 
not exceed eight times the thickness of the slab, nor 
one-half the clear distance to the next beam. 


‘For beams having’a flange on one side only, the effective flange 
width to be used in design shall not exceed one-tenth (7'y) of the 
span length of the beam, and its overhanging width from the face of 
the web shall not exceed six (6) times the thickness of the slab 
nor one-half (4) the clear distance of the next beam.” 

This practice is conservative. (See tests, pp. 36 to 38.) 

Isolated beams in which the T-form is used only for the purpose 
of providing additional compression area of concrete should prefer- 
ably have a width of flange not more than four times the width of 
the stem and a thickness of flange not less than one-half the width 
of the web of the beam. 

Width of Flange Required by Different Building Codes.—In 
large cities the Building Codes must be followed. The dimensions 
of the flange required by the Codes of the Cities of Boston, Chicago, 
Cleveland, New York, and Philadelphia are given in Fig. 72, p. 219. 

Cross Section of Web as Determined by the Diagonal Tension.— 
The width of the web of a T-beam is governed by the shearing 
stress (see p. 244) and by the layout of the tension bars (see p. Zia) 

The area of the web required for shear involving diagonal ten- 
sion, using notation on p. 215 and letting V = total external shear, 
and v = shearing unit stress, may be found from the formula (see 
also Formula (37), p. 247). 

Area of Web Required by Diagonal Tension.— 


v(d ~ 5) oe 4 (approx: fer eee) 
or 


e|< 


Wid S 71 ee a 


This means that the area of web at any point in the beam (con- 
sidering this up to the middle of the slab) must not be less than the 


DESIGN OF T-BEAM 219 


total shear divided by the maximum allowable unit shear for the 
reinforced beam. | 

The vertical unit shearing stress, v (used as measure of diagonal 
tension), in a beam effectively reinforced with bent bars or stirrups, 
or both, should be limited to 120 lb. per sq. in. for ordinary con- 
crete having a compressive strength (in cylinders) of 2000 lb. per 
sq. in. at twenty-eight days. 


Ce 
Boston & Cleveland \~«----- Gi-----: --b!--3 ceaOberan == b not to exceed d l 
New York & Tulare a a Gt ----->p<-- b'-->he---- 6t-------> 
@ UMax. + L Max. 
Chicago ~ <---------- 4 Spacing of beams------------ b not to exceed i l 


: l =Span of beam 
Fig. 72.—Width of Flange Required by Building Codes. (See p. 218.) 


Minimum Depth of T-Beam at Center of Span.—A minimum 
depth is the depth at which concrete and steel are stressed simul- 
taneously to their working limits. It is governed by the law that 
the compression in the flange must not exceed the working com- 
pressive stress in the concrete. Greater depth than the minimum, 
with proportionally decreased amount of steel, is generally used 
for economy. A smaller depth would give excessive compressive 
stresses. 

To find the minimum depth, the rectangular beam formula (1) 
p. 204, may be used where the depth of the beam is not greater than 
four times the thickness of the slab, the breadth of the flange of the 
T-beam being used in this formula for the breadth of the rectangular 
beam. For ratios of depth of T-beam to thickness of slab larger 
than four, the rectangular beam formula gives unsafe results and 
the following formula should be used: 


220 REINFORCED CONCRETE DESIGN 


Minimum Depth of T-beam at Center of Span, 


M 


In the above formula, Cy is a constant from pp. 894 and 895, 
depending upon the stresses f, and f, and the ratio _ Since the 


depth of beam, d, is not yet known, the depth required by shear may 
be taken as a close approximation, in computing the ratio of > The 
use of the formula is illustrated in the example on p. 896. 
Economical Depth for T-beam.—For T-beams, a depth greater 
than the minimum depth required by compression stresses will 
usually be found economical. By increasing the depth of the stem, 
the area of the tensile reinforcement is decreased, with a consequent 
reduction in cost of steel. The volume of concrete in the web, and 
as a consequence the cost of concrete, is increased; but, up to a cer- 
tain point, the reduction in the cost of steel is larger than the increase 
in the cost of concrete, and the cost of the beam as a whole is there- 
fore reduced. The most economical depth is the depth for which the 
‘nerease in cost of steel is equal to the decrease in cost of concrete, 
but beyond which any further increase in the depth would cause an 
increase in cost of concrete larger than reduction in cost of steel. 
The following convenient formula for the economical depth of 
the beam was developed by Professors Turneaure and Maurer? 
Using the notation given p. 215 and 


y = ratio of cost of cubic foot of steel in place to cost of 
cubic foot of concrete in place. 


Economical Depth of Beam, 

rM  t oe 
ay + (24) 
From this formula the economical depth for different widths of stem 
may be computed, and the most suitable combination of depth of 


beam and width of stem may then be selected. 
The value of the ratio r ranges from 35 to 75. It is computed 


Aig 


3 Turneaure and Maurer, “Principles of the Reinforced Concrete Construc- 
tion, Second Edition, p. 238. 


DESIGN OF T-BEAM 221 


as follows: If concrete in place costs 40 cents per cubic foot and steel 
in place 4 cents per pound or 4 X 490 = 1960 cents per cubic foot, 
the ratio is r = +38° = 49. 

In continuous beams add from 25 to 40 per cent to the unit cost of 
steel to allow for the overlapping of bars at the supports. 

Value r as Affected by Cost of Formwork.—In computing the 
value r above, only cost of materials was considered because the 
actual cost of formwork is little affected by small changes in depth 
of beam. The supports for the forms and the number of joints 
remain the same and the labor of making, erecting and stripping 
the form also is little affected by changes of depth of beam. The 
only effect caused by changes in depth is a small difference in cost of 
lumber for the sides, which is negligible, specially when forms are 
reused several times. 

Often, however, in practice the cost of formwork is computed by 
multiplying the area of the bottom and the sides of the beam by a 
certain unit cost. In such case the computed cost of formwork is 
affected appreciably by changes in depth of beam. The economical 
depth of beam, when estimated on that basis, may be obtained by 
Formula (24), p. 220, in which value of r is found as follows: Com- 
pute volume and cost of concrete for a section of beam, 1 ft. long and 
1 ft. deep with width of stem equal to 6’. Add to this the cost of 
2 sq. ft. of formwork based on the units used in estimating. Divide 
this total cost by the volume of concrete previously computed. 
Substitute this new unit cost for the cost per cubic foot of concrete 
in figuring the value 7. . 

Area of Steel in a T-beam.—The area of cross section of steel 
in tension may be obtained very closely by the following formula: 


' Required Area of Steel, 
M 


A, an rar 
jd 
The value of 7 may be taken from the following table: 


(25) 


Value of j for Different Ratios of Depth of Beam to Thickness of Slab : 


a i 
= ee ae 0.12 0.14 0.16 0.18 0.20 0.22 0.24 0.26 0.28 0.30 
» Poel 0.95 0.94 0.98 0.98 0.92 0.91 0.90 0.90 0.89 0.88 0.88 


Sometimes jd is assumed to equal 2d, or d — but the error is 


9? 
larger than in the above table. 


222 REINFORCED CONCRETE DESIGN 


More exact values of 7 may be found from the table on p. 897, 
or by the use of formulas on p. 184. 

Minimum Depth at Support in Continuous Beams.—At the sup- 
port of a continuous beam, owing to the change of the bending 
moment from positive to negative, the compression flange of the 
T-beam is in the tensile zone and is therefore ineffective. ‘The com- 
pression zone consists only of the width of the stem, and the T-beam 
changes into a rectangular beam with a depth equal to the depth of 
the beam and a width equal to the width of the stem. The com- 
pression stresses, therefore, are much larger than in the center and 
may require special provision. 

In all cases, some of the reinforcement used in the center for 
positive bending moment is carried straight at the bottom, through- 
out the full length of the beam. ‘This reinforcement, if carried a 
sufficient distance into the column to develop the required com- 
pression stresses, may be utilized as compression reinforcement. 
The beam then becomes a rectangular beam with steel in top and 
bottom. 

On account of the great reduction in the compressive area, the 
compression stresses at the support may be excessive and the con- 
dition at the support may determine the depth of the beam. The 
required depth at the support will depend upon the amount of com- 
pression reinforcement used. Usually, about one-half of the bottom 
steel is available as compression reinforcement. When the required 
area of tensile steel is the same at the support as in the center and 
one-half is bent up and carried over the support, the ratio of the 
available compression steel to the tensile steel at the support is oe 
This is the most desirable ratio to use. If this ratio is not sufficient, 
additional compression steel may be obtained by extending the 
bottom steel from the adjoining span a sufficient length to make it 
available as compression reinforcement in the span under consider- 
ation. With such design the ratio of compression steel to tensile 
steel may be increased to one. Larger ratios than one can be obtained 
only by introducing additional short bars at the bottom. 

After the ratio of the area of compressive steel to the area of 
tensile steel is selected, the required minimum depth of beam may 
be found from the following formula: 

Minimum Depth at Support, 


M . 
d — 1.054). o (leu, evoke See oFeRes (26) 


DESIGN OF T-BEAM 223 


In this formula, M is bending moment at support, b and d are 
dimensions of rectangular beam, f; is tensile stress in steel, and p; 


; Oe : 
is the tensile steel ratio, bd? the magnitude of which depends upon 


the accepted compression stresses in concrete and also upon the 
ratio of the area of the compression steel to that of the tensile steel. 
The value of p; may be taken from pp. 904 to 907. The use of the 
formula is illustrated by the example given below. The allowable 
compression stress in concrete at the support is larger than in the 
center of the beam, as explained on p. 282. 


Example 3.—Given the bending moment at support, M = 1 200 000 in.-lb.; 
allowable working stresses, fe = 750, fs = 16000, n = 15; width of stem, 
6 = 12 in. 

Find minimum depth if it is desired to limit the area of compression steel to 
one-half the area of tensile steel, so that - = 0.5. 

i 

Solution.—Assume the ratio of depth of compression steel to depth of beam, 
a=0.1. From Diagram, p. 904, in the section for f, = 750 and f,; = 16 000, 
and in the column for a = 0.1, we find by interpolation that a value of p’ = 0.006 
corresponds to a value of pi = 0.013. This gives substantially the desired ratio 
between the compression steel and the tensile steel. The value of p1 = 0.013 
will be used in Formula (26), p. 222. 

Thus, minimum depth 


—— ey 


d= 1.054|-™ ~ 1.054] mg O00 = 1.05V480 = 23 i 
eee NS pee N12 «0.018 X 16.000,” ari ea 








Details of Design.—In designing T-beams, diagonal tension must 
be investigated and web reinforcement provided where necessary 
(see p. 247). Bond stresses must be computed (see p. 262). In 
continuous beams, negative reinforcement must be provided (see 
p. 281). 

To insure proper T-beam action and prevent cracks through the 
slab adjacent to the stem, reinforcement at right angles to the beam 
or girder is necessary. When the principal slab reinforcement is 
parallel to the girder, short bars should be placed at the top of the 
slab, transversely over the girder and extending on both sides well 
into the slab. The ratio of the area of this reinforcement to the 
cross section of the slab should be at least 0.003. 

The example on p. 578 illustrates the use of the formulas and 
principles of design. The selection of bending moments is treated 
on p. 275. 


224 REINFORCED CONCRETE DESIGN 


To Find Stresses f, and f, in T-beam.—TIf the dimensions of beam, 
the amount of steel and the bending moment are known, the stresses 
may be found as follows: 

Find k and j from Formulas (19) and (20), p. 134. 

Find f, and f, from Formulas (22) and (23), p. 134. 

The work may be simplified by use of the table on p. 896, giving 


values of k, j andC,. Compute ratios ae and M For these find k 


from the Table of k. Knowing k and a find value of j from the Table 


of j. The stress in steel is then founed from 


M 
Je = Asjd 
The stress in concrete is found from 


fot A, = Crh PUT Sawet (27) 
where C7 may be taken from table on page 896. 

T-beam Design Considering Compression below Flange.—In a 
large T-beam, it is often desirable to take advantage of the com- 
pression stresses in the web below the flange, especially when the 
compression stresses, as computed by the approximate formulas, 
are excessive and it would be necessary to increase the compression 
area. Computed by the formulas that take account of the com- 
pression below the flange, the compression stresses are smaller. 
In instances where the thickness of the flange depends upon com- 
pression stresses in the T-beam, the use of these formulas reduces © 
the amount of concrete in the flange and thereby the dead load. 

The formulas on-p. 137 are too complicated and may be replaced 
by the following simple method, which gives equally exact results. 

Simplified Method.t—This method, which gives exact results, 
need be used only when the compression stresses in the beam are 
practically equal to the maximum allowable stress. Where the com- 
pression area is in excess, the approximate formulas on p. 134 give 
satisfactory results. The simplified method is most useful in large 
T-beams, in isolated T-beams, and for girders in joist-constructed 
floors. Its use frequently saves appreciable amount of concrete. 

Consider a T-beam section, as shown in Fig. 73,.p. 225. If the 
stress in concrete in such a beam equals the maximum allowable, fe, 

4'This method was devised by Edward Smulski. 


DESIGN OF T-BEAM 220 


and the stress in steel is fs, the position of the neutral axis of the 
T-beam is located in the same place as in a rectangular beam of the 
same depth and with the same unit stresses. 

The T-beam may be considered as consisting of two parts: (1) 
a rectangular beam with a width equal to the width of stem, 0’, 
and a depth equal to the depth, d, of the original beam; and (2) 
an auxiliary T-beam, the width of flange of which is equal to the sum 
of the projections of the T-beam and the depth equal to the depth 
of the original T-beam. The dimensions of each part and the 
area of steel required for each may be determined separately and 
the results added. The rectangular beam may be designed by 





Rectangular Beam Auxiliary T-Beam 


Fig. 73.—Illustration of Simplified Method of T-Beam Design. (See p. 224.) 


rectangular beam formulas. In the auxiliary T-beam there is on 
concrete below the flange; therefore, the formulas on p. 216 give 
exact results. 

This method is particularly advantageous if it is necessary to 
provide compression reinforcement in the T-beam, as may be the 
case in theater construction, where not only the depth of the beam 
but also the width of the flange is limited. In such cases, the T-beam 
is considered as composed of a rectangular beam with double rein- 
forcement and an auxiliary T-beam without concrete below the flange, 
and the results are combined. 

Case 1.—The procedure with ordinary beam and slab may be 
seen from the following examples: 


226 REINFORCED CONCRETE DESIGN 


Example 4.—Design a T-beam for an external shear, V = 134 000 Ib. anda 
bending moment M = 16 000 000 in.-lb., when the depth is limited to 60 in. and 
the thickness of slab is 5 in. The allowable stresses are fe = 650, fs = 16 000, 
i =oela. 

Solution.—The required width of stem for shear will be determined first. 
Since h = 60 in., d = 60 — 5 = 55 in, andv = 120 and, assuming j = 0.9, the 
required width of stem is 

134 000 


~ 0.9X 55 xX 120 


This value of b’ is accepted and the cross section of the T-beam is then as shown 
in Fig. 74, below. 


b’ = 92.6 in.  Usebenz3 1a 






! 
! 
‘ 
I 
| 
! 
j 
| 
| 
! 


tee ecceee 
phos Saas I “ye eoccce! : 


Loge Aar= 78d=9.9 89: In: ‘Agg = 12.0 89. In 





T-Beam Rectangular Beam Auxiliary T - Beam. 
Balanced Design ; 


Fic. 74.—Cross Section of T-Beam and Auxiliary Beams. (See p. 226.) 


Divided into two parts, the beam consists of a rectangular beam with 

b = 23 in., h = 60 in., d = 55 in, and an auxiliary T-beam with b = 80 in., 

= 5 in., and d = 55 in. 

For the specified stresses, for rectangular beam, the ratio of steel isp = 0.0077, | 
andj = 0.874. The corresponding area of steel is 

Ag, = 0.0077 X 23 X 55 = 9.7 sq. in. 
The moment of resistance of the rectangular beam, from formula M = idAsfs, iS 
M = 0.874 X 55X 9.7 X 16 000 = 7 462 000 inh. 


The auxiliary T-beam must resist the difference between the total bending 
moment and the moment resisted by the rectangular beam, or 


17 500 000 — 7 462 000 = 10 038 000 in.-Ib. : 
For d = 55 in., t = 5 in, the ratio of thickness of slab to depth is 


b ty Pang Bo 
Tyee a 


for which, from table on p. 221, 4.=.0;,95. » The required area of steel, from 
formula As = af is 
jdfs 10 038 000 


4. De ae ee 
>= 9 95 X 55 X 16 000 al 


DESIGN OF T-BEAM 227 


To determine whether compression stresses are satisfactory, compare the area of 
steel, just computed, with the limiting area of stecl. The limiting area of steel 
corresponding to the allowable working stresses, from Formula (19a), p. 216, is 


As = bdpm = 80 X 55 X 0.0032 = 14.1 sq. in., 
where p» is taken from diagram, p. 894, corresponding to the stresses fe = 650, 
t 
fs = 16 000, n = 15, and NT 0.09. This is larger than the value found above; 


therefore, the design is satisfactory, as far as compression stresses are concerned. 
The total area of steel in the T-beam is 


A, = Asi + Ay ee ++ Teer? le7 Sq. in. 


Comparison of Result of Exact Formula and Approximate For- 
mula.—To get a comparison between the areas of steel and com- 
pression stresses computed in the previous example by the exact 
(simplified) method and by the approximate method, it is now 
necessary to compute the steel areas and compression stresses by 
the approximate method. 

Steel Areas.—If the compression stresses below the flange are 
not considered, the area of steel, from Formula (19), p. 216, would 


be 
17 500 00 


= ron x ba x16. 000 = 


A, 
It will be noticed that this area of tension reinforcement is smaller 
than the area A; = 21.7 sq. in., obtained previously by the exact 
(simplified) formula. This is easily explained by the fact that in. 
the approximate formula the moment arm, jd = 0.95d was used for 
the total bending moment, while in the exact formula the moment 
arm jd = 0.95d was used for the T-beam part and jd = 0.874d for 
the rectangular beam part of the T-beam. 
Compression Stresses.—The compression stresses, for the case 
where compression in the stem below flange is not considered, will 
be found from formula f, = C7f,;. The value of Cy will be taken 


from Table 13, p. 896, for corresponding In the present case, 





eA nla X 20.9. ae hee , is 
ee maven 0.6 and aoa 0.09, for which k = 0.4 
and Cp = 0.044. 


The figured stress in concrete, therefore, would be f,; = 0.044 X< 
16 000 = 704 Ib. per sq. in. This stress is about 8 per cent larger 
than the allowable stress, f. = 650, and to reduce the stresses to the 


228 REINFORCED CONCRETE DESIGN 


allowable working stress it would be necessary either to increase 
the depth of the slab (since the depth of beam is limited) or to add 
compression reinforcement. By the use of the exact formula, how- 
ever, which correctly utilizes the compressive resistance below the 
flange, this necessity is avoided and an appreciable saving made in 
materials. 

Case 2.—Another use for this (simplified) method is for deter- 
mining the required flange area for a beam where the flange is pro- 
vided solely to increase the compression area of the beam. Such is 
the case in beams carrying joist construction, where the floor is 
thickened on both sides of the beam, and in isolated beams with 
projecting flanges. This method is not restricted to cases where 
the flanges are of uniform thickness, but may also be used with 
unusual flanges, as shown in Fig. 75, below. 





Lyte] As" Asr*Asz tAss|. ys{As1= ped ‘As2 As3-” 


T-Beam. Rectangular Beam : Auxiliary T - Beams 
Balanced Design 





Fic. 75.—T-Beam with Unusual Flange. (See p. 228.) 


In such cases, the T-beam is replaced by a rectangular beam and 
two T-beams, one of which has the thicker flange and the other the 
thinner flange. The beams are shown in Fig. 75. 


Example 5.—Design a beam to support joist construction consisting of joists 
8 in. deep, for which external shear, V = 100000 lb., and bending moment, 
M = 15000000 in.-lb. The total depth of beam is limited to 54 in. The 
allowable stresses are fe = 650, fe = 16 000, n = 15,9 = 120. 

Solution.—Total depth of beam, h = 54 in., allowing 2 in. for fire proofing 
and assuming three layers of one-inch bars with one-inch separators between 
layers. The distance from bottom to center of steel is 2 + 2.5 =4.5in. The 
effective depth of beam is d = 54 — 4.5 = 49.5 in. 

For V = 100 000 lb., the width of beam required for shear is 


Va 100 000 
“0 874 < 49-5 120 





b = = 19.2in.:, Use 20 m: 


DESIGN OF T-BEAM 229 


For M = 15 000 000 in.-lb. and the depth d = 49.5 in., the required width 
of a rectangular beam, from Formula 2, p. 204, is 


M 15 000 000 


b = oh ee tet mantic 
Rd? 106 X 49.52 





- in. 

This width required by bending moment is much larger than the width 
required by shear. A rectangular beam would be too expensive and too heavy. 
It will be cheaper to use a T-shaped beam in which the width of the stem is equal 
to the width required by shear and compression is supplied by a flange. 

The required dimensions of this T-beam will be found by considering the 
compression below the flange. The simplified method will be used, assuming 
the T-beam replaced by a rectangular beam and an auxiliary T-beam consisting 
of the two projecting flanges on each side of the stem only. 

The dimensions of the rectangular beam are already known. They are: 
b = 20 in., d = 49.5 in. (6 of rectangular beam becomes 6’). For f, = 650, 
fs = 16 000, the ratio of steel is p = 0.0077 and 7 = 0.874 (see table, p. 205). 
The area of steel for the balanced rectangular beam, from formula A, = pbd, is 


map = 00017 X 4075 xX 20 = 7°6 sq. in. 
The moment of resistance of the beam, from M = A,jdf,, is 
M, =7.6 X 0.874 X 49.5 X 16 000 = 5 030 000 in.-lb. 


The bending moment to be resisted by the auxiliary T-beam is obtained by 
subtracting from the total bending moment the moment of resistance of the 
rectangular beam, : 


M, = 15000 000 — 5 030 000 = 9 970 000 in.-lb. 


It is now necessary to decide upon the thickness and the width of the flange. 
To simplify the formwork, a thickness of flange equal to the depth of the joist 
will be accepted. Therefore, ¢ = 8 in. The width (6 — 0b’) will be computed. 


t 
For d = 49.5 in. and t = 8 in., the ratio Ee 0.16, for which, from table on 


) M 
p. 221,7 = 0.93. ‘The area of steel, from A, = —, is 
jdfs 
9 970 000 


hi Sei a 
* 0.93 X 49.5 X 16 000 


= 13.5 sq. in. 


The width of flanges of the auxiliary T-beam is obtained from the fact that 
the above area of steel is equal to the maximum allowable steel area permitted 
by compression stresses. 

The equation for maximum allowable steel area in a T-beam is As. = bdpm, 


: t 
in which pm is a constant obtained from diagram p. 894, for known f;, fs, and 7 


In this case, pm = 0.0051 and the steel area A,z = (6b — b’) X 49.5 X 0.0051. 


230 REINFORCED CONCRETE DESIGN 


Equating this to the steel area computed above, 13.2 = (b — b’) X 49.5 X 0.005, 
from which 
13.5 
b— db! = —— = 53 in. 
0.253 
To get the total width of flange, it is necessary to add to the above value the 
width of stem b’ = 20 in. Hence b = 52 + 20 = 72 in. 
Final Design, 
Concrete dimensions, b = 72 in.; 


b=" 20 im.; 
a’ =49.5 ins 
h = 54. 
Area of steel A =Agi t+ Ag = 7.6 + 12.9 = 20.5 8q. in. 


T-BEAM WITH COMPRESSION REINFORCEMENT 


Use of Compression Reinforcement in T-beams.—Compression 
stresses are seldom a controlling factor in T-beam design. Occa- 
sionally, however, the stresses may be larger than can be taken care 
of by the concrete, and compression reinforcement may then be 
advantageous. | 

Compression reinforcement is required when not only the depth 
of the beam but also the available width for the flange is limited. 
One such case is that of beams supporting balconies in theater con- 
struction. | 

Compression steel may be useful also in a series of T-beams of 
similar design, of which one beam carries a larger load than the 
others. If the compression stresses in all but the heavy beam are 
satisfactory, it is cheaper to add. compression steel rather than to 
change the concrete dimensions. 

Formulas to Use.—In designing T-beams requiring compression 
reinforcement, formulas taking into consideration the compression 
in the stem below the flange should be used. The simplified method, 
in which the T-beam is considered as a combination of a rectan- 
cular beam and an auxiliary T-beam, may be used to advantage. 
The rectangular beam will be one with compression reinforcement. 

Problem to be Solved.—The problem in this connection usually is: 
Given the depth of beam, d, and the width and thickness of flange, 
b and t, and the width of stem, 6’. Determine the required tensile 
and compression reinforcement for a given bending moment, M, 
and specified unit stresses, fs and fo. (See Fig. 76, p. 231.) 

The problem is solved by computing the moments of resistance, 
M,, of a balanced rectangular beam without compression rein- 


T-BEAM WITH COMPRESSION REINFORCEMENT | 231 


forcement. For the specified stresses, the steel ratio, p, is taken from 
the table on p. 880. 

Then the area of tensile reinforcement equals As; = pb/d, and 
the moment of resistance is My = Asijdfs. 

Next, the moment of resistance, M2, is computed for the aux- 
iliary T-beam. ‘The tensile area is 


A52 = Pmbd, 


Compression Steel Ag 










Tae 


a “Vv -~Tension Steel 
o , 1 A,=A,, + Ago t Ags 
hese 


T - Beam with Compression Reinforcement 


| 





= AR eer 
arr RK baa shir 
| ' ‘ ' 
a Vibe tices 
} 8 1 bo 
! ! i! a=: 
an he 
Luo aL Leet 
ee Se Ae (see f 
Aggy 
Rectangular Beam Auxiliary T-Beam Steel Resisting Couple 


Balanced Design 


Elements Replacing T - Beam with Compression Reinforcement 


Fic. 76.—T-Beam with Compression Reinforcement. (See p. 230.) 


where pm is taken from diagrams 3 or 5 for the specified stresses and 


the proper value of . The moment of resistance 1s Mz = A.ojidf., 


where 7; 1s value from Table 13, p. 897. 

Finally, the bending moment to be resisted by the compression 
steel is equal to the difference between the total bending moment 
to be resisted, M, and the sum of the bending moments, M; and Mo, 
resisted by the rectangular beam, and the auxiliary T-beam, respect- 
ively. It is, therefore, M3 = M — (M; + Mo). 


232 REINFORCED CONCRETE DESIGN 


This bending moment may be considered as resisted by a couple, 
composed of compression resisted by compression steel and tension 
resisted by tensile steel. The moment of this couple is 


Mz = A,3d(1 — a), where d(1 — a) is the moment arm. 








From this : 
ee _ M — (Mi + M2) 
Asay 9) oT dl (28) 
The total amount of tensile steel is 
A, = Aa + Ait Ale pt (29) 
The compression reinforcement may be found from formula 
Teil 1—k 
Alm Mesias) tile i eee 


where k is the ratio of neutral axis and a the ratio of depth of com- 
pression steel. 


RECTANGULAR BEAM WITH COMPRESSION REINFORCEMENT 


Use of Compression Reinforcement.—When, in a beam of fixed 
concrete dimensions, the compression stresses in the extreme fiber 
are larger than the allowable unit stresses, the beam may be strength- 
ened by the introduction of reinforcement in the compression zone. 
This compression reinforcement resists fifteen times the stresses 
(for n = 15) which were resisted by the replaced concrete and thus 
reduces the stresses in concrete. To be effective, the bars composing 
the compression reinforcement must be straight and they must 
extend a sufficient distance beyond the support to transmit the 
stresses from the bar to the support. 


PROBLEMS IN CONNECTION WITH CoMPRESSION REINFORCEMENT 


The following problems may occur: 

Problem 1.—Concrete cross section fixed. Determine area of 
tension and compression reinforcement for given bending moment 
and specified working stresses. 

Problem 2.—Ratio of area of compression steel to area of tension 
steel fixed. Find required dimensions of beam for given bending 
moment and specified working stresses. 


RECTANGULAR BEAM WITH COMPRESSION STEEL 233 


Problem 3.—Concrete dimensions and areas of steel given. 
Find stresses in steel and concrete for any given bending moment. 

Problem 4.—Concrete dimensions and areas of steel given. Find 
moment of resistance of beam, and from this the load the beam is 
capable of carrying for specified working stresses. 

Solution of the Problems.—All these problems may be solved by 
using formulas given in the section on theory (see p. 137). These 
formulas, however, are too complicated. 

Easier Solutions are Provided by Two Simplified Methods, 
Devised by Mr. Edward Smulski and Given Below.—These methods 
give exact results, require less work, and are easier to understand 
and to remember than the elaborate formulas. 

By the first method, the required areas of compression steel may 
be obtained directly. Problem 1 above may be easily solved by 
this method. The method is especially valuable when handbooks 
(such as this volume) with tables of constants are not available. 

The second method, a modification of the first method, gives 
the required ratio of compression steel, p’, when the computed 
ratio of tensile steel, p1, exceeds the value, p, permissible for a simple 
beam; i.e., when the area of concrete is too small to resist safely 
all of the compression. This method may be used for solving Prob- 
lems 1 and 2, either directly or in conjunction with diagrams, p. 904. 
It is particularly useful for determining compression steel needed 
for negative moment in a continuous beam. 

Problems 3 and 4 are best solved by using the diagrams, pp. 908 
and 909, as explained on p. 228 and illustrated by examples opposite 

each diagram. 


Let A, = area of cross section of tensile steel in beam under 
consideration ; 
Asi = pbd = area ® of tensile steel in balanced: simple beam 
in which steel and concrete are stressed to specified 
allowable unit stresses, f, and f.; 
p = steel ratio for balanced design of beam with tensile 
steel only; 
= ratio, in beams with stecl in top and bottom, of total 
cross section of steel in tension to cross section of 
beam, bd; ) 


as 


5 The limiting value for which no compression steel is required. 


234 REINFORCED CONCRETE DESIGN 


p’ = ratio of cross section of steel in compression to cross 
section of beam, bd; 
k = ratio of depth of neutral axis to depth of beam, Jie 
a = ratio of depth of compressive steel to depth of beam; 
Awe = A, — As = extra tensile steel, in beam with steel in 
top. and bottom, to be balanced by compression 
steel ; 

A’, = area of compression steel required to balance extra 
tensile steel = p’bd; 

M = moment of resistance or bending moment for beam 
with steel in top and bottom; 

M, = moment of resistance of beam without compressive 


reinforcement. 


First Method. Simplified Method of Design of Beam with Steel 
in Top and Bottom.—As evident from Fig. 77, the position of the 
neutral axis for a beam with steel in top and bottom is the same 
as for a rectangular beam of the same dimensions and subjected to 
the same stresses, f, and fs. The beam with steel in top and bottom 
may, therefore, be considered as a combination of (1) a rectangular 
beam without compression steel and (2) a couple of equal forces, one 
of which forces is the compression resisted by the compression steel 
and is equal to A’s f’s, while the other force is the tension resisted by 
tensile steel and is equal to Asofs. 

The moment of resistance, M, of a beam with steel in top and 
bottom may therefore be separated into two moments: (1) the 
moment of resistance, Mj, of a rectangular section of balanced 
design with bottom reinforcement only, and (2) Me, the moment 
of resistance of the couple of forces. By a beam of balanced design 
is understood a beam with an area of tensile reinforcement, As = pbd, 
such that both the maximum tensile and the maximum compression 
stresses are reached at the same time. 

The moment of resistance of a simple beam of balanced design is 


M, = f:Asjd in which Asi = pbd. 


The value, p, is a definite ratio depending upon the specified values 
Ofte a... a0 J 

The moment of resistance of the couple of forces is obtained by 
multiplying the tensile force, As2fs, by the moment arm d(1 — a), 


Ms: — Af dl = a). 


RECTANGULAR BEAM WITH COMPRESSION STEEL 235 


The value of M; is fixed for any size of beam and specified stresses 
by the strength of the concrete. The value of As; is also fixed. 

The value of Mz depends only upon the tensile area As2 and the 
corresponding compression area A’;. Within practical limits it may 
be made of any magnitude. 


The moment of resistance of a beam with steel in top and bottom 
is M = M, + Mz. 

















t 
] Res 
a: Compression Steel A, wins - f- ; 
| \ ro loo Cc _ Compression 
Taare ed eee = in Steel 
hs | 
Potts Neutral ~-Total 
eae 3 lage \ Compression 
LS WNW X1S \ : 
Sere ‘Comp. in Concrete 
| | 
2 Arr eA AE 


es --Tension Steel 
-- b-->| A,= pbdtA,, 


Beam with Steel in Top and Bottom 


d(1-k) 

















Neutral = 
Axis ea 
~~ 
d (1 - k) rs 
| 
Benin steel) ee Na- He in 
podf, ee 
: Ago f, 
Rectangular Beam Balanced Design Steel Resisting Couple 


Elements Replacing Beam with Steel in Top and Bottom 
Fic. 77.—Beam with Steel in Top and Bottom. (See p. 234.) 


The above relation may be used for solving Problem 1, in which 
the dimensions, b and d, are given and it is desired to find the areas 
of tensile and compression reinforcement to resist a bending moment, 
M, with specified stresses fs, fc, and n. 

The ratio of steel, p, corresponding to the specified stresses, is 
found first, either from table on p. 880 or from Formula (5), p. 130. . 
(For instance, for f; = 16000, f. = 800, n = 15, the steel ratio is 
p = 0.0107.) The area of steel is Asi = pbd and the moment of 
resistance of a balanced rectangular beam without compression 
reinforcement is 


M, = jdAaf, = jdpbdf,. 


236 REINFORCED CONCRETE DESIGN | 


Subtract this value from M: The remainder, M — Mi, must be 
resisted by the steel couple whose moment of resistance 1s 


Mos = Asefd aa a). 
Thus, A,fd(1 —a) =M—M, and 
M— M, 


The total area of tensile steel is 
2 oe M—M, 
A, = AA -- Ag — pbd ct. aera este (32) 
The area of compression steel is 
Be (Neco x < 
Aa Aw ee (33) 


This method is exact. It does not require any tables. The 
values of p and k corresponding to the specified stresses are the only 
constants required. 

Second Method. Determining Ratio of Compression Steel, p.— 
The method given below may be used for determining the required 
ratio of compression steel, p’, if the ratio of tensile steel, 1, is larger 
than the allowable ratio, p, for a balanced simple beam. ‘This is 
a modification of the previous method. It is particularly useful in 
continuous beams at the support. 

For any accepted stresses, fc and fs, and the ratio n, the location 
of the neutral axis is the same for beams with compression rein- 
forcement as for simple beams. 

The formula for tensile steel ratio for beam with compression 
reinforcement is (see Formula 41, p. 189). 

ke? k—- a 
aso 2n(1 — k) TP ocae 
° ke Pweg : : 
The first term in this formula, nd — is identical with 


2 
Formula 6, p. 130, p = mF in which p is the ratio of tensile 


* The formula for the area of compression steel is obtained from the require- — 
ment that the total tensile stresses in the couple of forces (see Fig. 77) must be 
equal to total compression stresses. The ratio of unit stresses in tension and 
compression is equal to the ratio of their distance from neutral axis. 


RECTANGULAR BEAM WITH COMPRESSION STEEL 230 


steel for a balanced simple beam. This may be substituted in the 
above equation and 


jk—a 
Php Palliy?.qeaty 





From this: | 
Required ratio of compression steel, 


ll (34) 





= (p 


This formula gives the ratio of required compression steel if the 
ratio of tensile steel, p:, necessary to resist a bending moment, is 
larger than the value of p corresponding to the specified stresses in a 
balanced beam without compression steel. 

For instance, assume that the computed ratio of tensile steel to 
resist bending moment is p; = 0.0125. If the allowable stresses are 
f. = 750, fs = 16 000, and n = 15, the allowable ratio without com- 
pression reinforcement is p = 0.0097, and value of k = 0.414. The 
required ratio of compression steel is then 


0.414 


= (0.0125 — 0. UND eres 


The value of a is the ratio of the distance of the center of com- 
pression steel from the surface to depth of beam. 

The amount of ee steel (and from this the value pj) is found 
from formula A, = = ip in which j is assumed as 0.9. This value of j 
is near enough for practical purposes. If desired, after computing 
the ratio of compression reinforcement, the correct ratio of j may be 
taken from the diagram on p. 910 ciate the values of pi and p’ revised 
accordingly. 

Solving Problems 1 to 4 (p. 232) by Means of Diagrams.— 
Formula (34) for the required compression steel was used in _pre- 
paring Diagrams, p. 904. From these the required ratio of com- 
pression steel may be taken directly for any ratio of tensile steel 
and different allowable stresses, f. and f,. 

Problem 1 may be solved by using Diagrams as follows: iki this 
problem it is required to find the amount of tensile and compression 
steel for given moment and stresses. For the given bending moment, 


the area of tensile steel is computed by formula A, (assuming 


M 
~ 0. 9df, 


238 REINFORCED CONCRETE DESIGN 


7 = 0.09). From this p: is found. The value a is estimated. 
Then the required ratio of compression steel is taken directly from 
the diagram corresponding to the specified stresses, fs and f-, and to 
the value a. Diagrams, p. 908, may also be used to solve the problem. 

Problem 2 also may be solved by means of diagrams. In this 


problem, dimensions of beam are required, with the ratio of 7 fixed 
1 


and stresses f. and f, specified. In solving, the value of a is estimated 
first. Then reference is made to the section for the specified stresses 
and the column for the nearest value of a. From this, a value of p’ 
is found which has the required relation to p:. For instance, if for 
f. = 16 000, f. = 750, n = 15, it is required to get a design in which 


/ 
ratio of aa 1, when a = 0.08. In the section for the specified 


f, and f, and on the curve for a = 0.08, it is found by inspection that 
a value of p’ = 0.016 corresponds to pi = 0.016. These values 
satisfy our requirement and may be used in design. (See example, 
p. 903.) 

For solving Problems 3 and 4, with given concrete dimensions 
and steel areas, where the relation between the stresses f. and f; 1s 
not known, the diagrams on p. 904 cannot be used advantageously. 
For this purpose, the diagrams on pp. 908 and 909 were prepared; 
from these the relation between the steel ratios, pi and p’, may be 


found for any ratio of unit stresses of ie and any value of a, and 


a 


15fe 
of steel ratios, pi and p’, and any value of a. 

In solving Problem 3, where it is required to determine the 
stresses for known dimensions of beam and known areas of steel, the 
values of p1, p’, and a are computed first. The tensile stresses may 


conversely the ratio of stresses may be found for any combination 


then be computed from the formula f; = pee (assuming j = 0.9). 


s 


The ratio of unit stresses, oe is found from the diagram for nearest 
value of a and for the known values of pi and p’. The tensile stress, 


fs, and the ratio, 7 being known, the compression stress is easily 





found by dividing the stress f, by 15 times the determined value for & 


RECTANGULAR BEAM WITH COMPRESSION STEEL 239 


Problem 4, in which moment of resistance is required for known 
dimensions and specified unit stresses, may be solved as follows: 
The moment of resistance may be governed either by the steel 
stresses or by the concrete stresses, whichever are smaller. The 
simplest method is to determine the moment of resistance governed 
by steel stresses and then find whether the compression stresses for 
this moment are within the specified working limits. The moment 
of resistance governed by steel may be found directly from formula 
M = 0.9dA,f., in which all values are known. (The value of moment 
arm is assumed j = 0.9.) To find compression stresses correspond- 
ing to this moment, compute, from known areas of steel and concrete, 
the steel ratios p; and p’ and ratioa. From the diagram for value of 
i corresponding to 
the steel ratios, p1 and p’. The compression stress in concrete 
may now be found by dividing the stress in steel, f,, by 15 times the 


a nearest to the computed value, find the ratio 


ratio ee, which was found from the diagram. If this compres- 


sion stress is smaller than the specified stress, the moment of resist- 
ance of the beam, as determined by tensile steel, is satisfactory. If 
the compression stress in concrete is larger than the allowable stress, 
it signifies that the moment of resistance of the beam is governed by 
compression stresses; hence the moment of resistance determined 
from tensile steel is too large, and the load the beam is able to carry 
is lighter than would follow from this moment of resistance. 

If the beam is already built, the moment of resistance governed 
by the concrete is computed, and from this is found the load the 
beam is able to carry. For this moment of resistance, the stress in 
if aie 
15, is 
already known. The reduced stress in steel may be computed by 
multiplying the allowable stress in concrete by n and by the ratio 

fs 
15f. 
is the moment of resistance as governed by compression stresses. 

If the beam is not built, the moment of resistance as governed by 
concrete should be increased by adding compression steel. Or, if 
the moment of resistance is sufficient, the amount of steel should be 





steel is smaller than the maximum allowable. The value 


The moment of resistance based on this reduced stress in steel 





reduced. The tensile stress may be found from the diagram oe 7 , by 


240 REINFORCED CONCRETE DESIGN 


substituting the value of f.. The value of the reduced f, then follows, 
and this will give the required moment of resistance. 

Moment Arm.—The magnitude of the moment arm for a beam 
with compression steel is different from the magnitude of the moment 
arm for a simple beam. Its exact value is 

Moment arm, Beam with Steel Top and Bottom, 


ae k op’ (k x —a 


This value may be obtained from the diagram on p. 910 for differ- 


ent ratios of stresses, z, and ratios of * In practice, 7 = 0.09 
aE 1 

may be accepted. After p’ and pi are computed, the value may be 

corrected by referring to the diagram. 


EXAMPLES OF BEAMS WITH STEEL IN TOP AND BOTTOM 


Example 6.—Given size of beam, b = 10 in. andd = 18 in. as determined by 
architectural requirements; bending moment, M = 500 000 in.-lb.; allowable 
unit stresses, fs = 16 000 Ib. per sq. in., fe = 700 1D. per eq. in., 7 = 16. Find 
required amount of tensile and compressive steel. 

Solution.—For f; = 16 000 lb.; f¢ = 750 lb. y trom table on p. 880, k = 0.414; 
j = 0.862; and p = 0.0097. The area of steel in a beam without compression 
steel is) As: = LOX 18 >< 0.0097 = 1.75 sq. in., and the moment of resistance 
is M, = jdAeifs = 0.862 X 18 X 1.75 X 16 000 = 435 000 in.-lb. By com- 
paring this moment with the bending moment of 500 000 in.-lb., we find a differ- 
ence of M — M1 = 500000 — 435 000 = 65 000 in.-lb. This must be provided 
for by extra tensile steel and by the corresponding amount of compressive steel. 
Since center of compression steel is distant ad = 2.25 in. from the top of beam, 
a = 0.126 and distance between tensile steel and compressive steel is 


d(l ua): S186 Ors ite 15.73 in. 


This is moment arm of couple composed of compression stress in steel and tensile 
stress of extra amount of tensile steel. Required extra amount of tensile steel, 
Ago, will be found by dividing, extra bending moment, M — M,, by dl — a)fs; 


65 000 
hence As: = = 0.26 sq..in. The total tensile steel therefore 
15.73 X 16 000 





equals 
As = Asi + Ag — 178 + 0.26 = 20 sq. in. 


From Formula (33), required amount of compression steel is 
—& 1 — 0.414 


se | 
tad If eer 126i 3 = 0.53 sq. in. 
gray (eam A as) has 





DIAGONAL TENSION 241 


Example 7.—Find required amount of compression steel for a T-beam at the 
support, where b’ = 12 in., d = 22 in., As = 4.0 sq. in. and the specified stresses 
are fs = 18000 and f, = 900, n = 15. Distance from top of beam to center of 


¢ 
. e 


compression steel, dj = 2} in. and a= = 0.102. 





Solution.—At the support, the T-beam changes into a rectangular beam 
(see p. 282). By dividing area of steel by b/d, ratio of tensile steel, 


4.0 


= ——— = ().0152. 
12 X 22 


Pi 


From table on p. 880, the ratio of steel for beams without compression steel, for 
stresses, fs = 18 000, and f, = 900, is p = 0.0107 and k = 0.429. Since ratio 
pi = 0.0152 is larger than p, compression steel is required. 

Ratio of required compression steel, from Formula (34), p. 237. 


— 0.429 0.571 


| ean 
’ = (0.0152 — 0.0107 = 0.0045 x —— = 0.0079. 
Beil * 0.825 ui 


0.429 — 0.104 


Therefore, an area of compression steel A’, = 0.0079 x 12 x 22 = 2.1 sq. in, 
is required to keep compressive stress in concrete within working limits of 900 lb. 
per sq. in., when the stress in steel is 16 000 lb. per sq. in. 


DIAGONAL TENSION 


Diagonal Tension Stresses.—In reinforced concrete members 
subjected to bending, diagonal tension stresses must always be com- 
puted, and if they exceed the allowable unit stresses for plain: con- 
crete, web reinforcement must be provided. 

All members subjected to bending develop, in addition to the 
tension and compression stresses due to the bending moment, ten- 
sion stresses in the web acting at an inclined plane and mainly 
caused by shear. In homogeneous beams, these stresses do not 
require any special attention. In reinforced concrete, on the other 
hand, they assume considerable importance. Full discussion of 
diagonal tension is given in the chapter on Theory, with which the 
designer should familiarize himself thoroughly. 

Formulas to be used in design and recommendations for their 
practical application are given below. 

Vertical Shearing Stresses as Measure of Diagonal Tension.— 
Concrete is strong in direct shear and capable of withstanding 
(in 2000-lb. concrete) a working shearing stress of at least 200 Ib. 
per sq. in. Concrete beams or slabs, therefore, always have suffi- 
cient area to withstand direct shear. 


242 REINFORCED CONCRETE DESIGN 


However, since shearing stresses have been accepted as a measure 
of diagonal tension stresses, and the resistance of concrete to diagonal 
tension is small, the unit shearing stresses must always be com- 
puted. If the stresses exceed the values for plain concrete, web 
reinforcement must be provided to take care of the excessive diagonal 
tension stresses. 

Usefulness of Web Reinforcement.—Numerous tests have 
demonstrated that a beam properly reinforced with stirrups or 
bent bars sustains three or four times as much load as the same 
beam without web reinforcement. 

This effectiveness in increasing the strength of the beam is 
unquestionable. There is a peculiarity in the action of the web 
reinforcement, however, particularly of the stirrups, which has 
sometimes created confusion and doubt—not as to their effectiveness 
for high load, but as to their usefulness in beams designed for ordi- 
nary working conditions. ‘Tests show that, until diagonal cracks 
appear, a reinforced concrete beam acts like a homogeneous beam. 
Moreover, the measured stresses in web reinforcement are small, 
and in some cases compression stresses, instead of tensile stresses, 
have been found in stirrups. The stirrups do not come into action, 
in fact, until after diagonal cracks develop. Since the allowable 
unit stress for diagonal tension is usually somewhat lower than the 
tensile strength of concrete, it can easily happen that the stresses in 
a beam will never be sufficient to crack the concrete and bring 
the stirrup into action. 

This fact, then raises a question as to whether there is any 
obiect in providing web reinforcement which may never be called 
upon to carry stresses. The answer is obvious. Reinforced con- 
erete beams are not designed for the condition at working loads, but 
for the condition at working loads multiplied by the factor of safety. 
(See also p. 125.) Web reinforcement should always be used, 
therefore, when unit stresses exceed the allowable stresses for plain 
concrete, irrespective of whether or not diagonal cracks are likely to 
occur under working loads. In beams without stirrups, final failure 
closely follows the appearance of the first crack; while with beams 
having web reinforcement, stirrups and bent bars represent a factor 
of safety which allows stressing of concrete in diagonal tension up 
to its ultimate strength. Under working loads, the stirrups may 
not act; but in case of overstressing, due to faulty construction or to 
occasional excessive loading, the stirrups prevent the failure of the 


DIAGONAL TENSION 243 


beam. The minute cracks that may develop are not dangerous 
and in many cases are hardly visible. They close after the excessive 
load‘is removed. 

Description of Web Reinforcement.—Web reinforcement may 
consist of (a) vertical or inclined stirrups; (6) bent-up tension bars; 
(c) a combination of stirrups and bent-up bars. 

Stirrups are steel members, usually U-shaped, made of bars of 
comparatively small diameter and independent of the main rein- 
forcement, extending from the tensile zone into the compression 
zone and anchored at both ends. Fig. 78, p. 243, shows different 
types of stirrups. 


ald to keep Stirrups in a _--Bars to keep Stirrups i. in Position.. 


Toor t 


Fig. 78.—Types of See (See p. 243.) 


Vertical Stirrups.-—Vertical stirrups are most commonly used. 
Although not placed in the theoretical direction of diagonal tension, 
they resist the stresses effectively. They are secure against slipping 
in a horizontal direction and are easy to keep in position during 
construction. As contrasted with bent-up bars, the number and 
spacing of stirrups can be varied to suit conditions. 

Inclined Stirrups.—Theoretically, the most appropriate rein- 
forcement to resist diagonal tension stresses would consist of inclined 
stirrups placed in the direction of the stress. The inclination of 
diagonal tension varies in different parts of the beam, being steepest 
near the end of the beam and flattening out toward the center of 
the beam, so that to follow the direction of the stress the inclination 
of the stirrups would have to be varied throughout the beam. This 
is impracticable. If used at all, the inclined stirrup should be 
inclined about 40° with the horizontal. The objection to inclined 
stirrups is that they are difficult to attach to the tensile reinforce- 
ment and to keep in place. Therefore, in spite of the superiority 
resulting from a closer coincidence between the direction of the 
stress and the direction of the bar, for practical reasons they are 
seldom used. 

Bent-up Bars.—Bent-up bars, if properly distributed and inclined, 
_ form excellent web reinforcement. Like inclined stirrups they have 
the advantage resulting from closer coincidence with the direction 


244 REINFORCED CONCRETE DESIGN 


of the diagonal tension stresses, and in addition they have the 
advantage of being rigidly connected with the main reinforcement. 

When bent horizontally at the top of the incline, and extended 
into the adjoining span or into the support, the bent bars have suf- 
ficient anchorage, so far as their function as web reinforcement 
is concerned. If not so extended, they must be provided in the 
compressive zone with hooks of proper dimensions to develop their 
tensile strength.- The only disadvantage of bent-up bars lies in the 
fact that their number in a beam is small and that the points of 
bending up may be governed by the bending moment requirements 
which, in turn, do not coincide with the requirements of diagonal 
tension reinforcement. 

Stirrups and Bent-wp Bars.—The most effective diagonal tension 
reinforcement consists of a combination of bent-up bars and prop- 
erly placed vertical stirrups. 

Anchorage of Web Reinforcement.—In the tensile zone of the 
beam, the web reinforcement must be continuous with the tensile 
reinforcement, as in bent-up bars; or, if stirrups are used, it is neces- 
sary to anchor them to the tensile bars by bending the stirrup around 
the lowest layer of tensile reinforcement in such a way as to develop 
the tensile value of the stirrup at the level of the tensile reinforce- 
ment. In the compression zone, the web reinforcement should be 
brought as near the compressed surface as possible. It should be 
provided with hooks of sufficient dimensions to develop, in con- 
nection with straight imbedment, the full tensile value of the bar in 
a distance 0.3d from the compression face of the beam. A semi- 
circular hook with a radius not less than four times the diameter of 
the bar will be found satisfactory. 

In continuous beams at the support, the tensile zone is at the 
top and the compression zone at the bottom of the beam. | 

Allowable Unit Stresses for Diagonal Tension.—The following 
unit stresses are recommended for concrete testing 2 000 Ib. per sq. in. 
at the age of twenty-eight days. This strength corresponds to the 
| : 2:4 mix used in general practice. 

(a) In beams without web reinforcement, the maximum unit 
shearing stress (being measure of diagonal tension) computed from 
- Formulas (36) and (37) must not exceed 40 Ib. per sq. in. 

(b) In beams provided with web reinforcement, the maximum 


unit shearing stress (being measure of diagonal tension) computed © , 


as before must not exceed 120 Ib. per sq. in. Not more than 40 lb. 


DIAGONAL TENSION 245 


per sq. in. of this stress may be considered as resisted by concrete, 
the balance being considered as resisted by web reinforcement. 

(c) In beams without web reinforcement, but where tensile 
reinforcement is anchored at the ends,® the maximum unit shearing 
stress, computed as before, may be 60 lb. per sq. in. 

(d) In beams with web reinforcement, where at least 50 per cent 
of tensile reinforcement extends the full length of the tensile zone 
and is anchored at the ends,° the maximum unit shearing stress 
may be increased to 180 lb. per sq. in. Not more than 60 lb. of 
this stress may be considered as resisted by concrete, and for the rest 
web reinforcement must be provided. The increased unit stress 
must not be used where too thin members would result. 

In continuous beams in the region of negative bending moment 
the top reinforcement must be anchored by extending beyond the 
point of inflection, to permit the increased unit stresses under (c) 
and (d). 

The 1924 Joint Committee allows a unit shearing stress of 240 Ib. 
per sq. in. for beams with end anchorage. This stress is considered 
by the authors as excessive, because it is likely to give thin members 
so filled with reinforcement as to interfere with proper placing of 
concrete. Moreover, such members would be particularly vulner- 
able in case of fire. 

The allowable unit stresses recommended above may _ be 
expressed in terms of the ultimate compressive strength of con- 
crete,’ f',, as follows: 40 lb. = 0.02 f’.; 60 Ib. = 0.03 f’.; 120 
lot 0.06.7 <5 180: 1b. = 0.09’; and. 240 Ib. = 0.12 f'.. Thus, 
for concrete of greater compressive strength, f’., than 2 000 Ib. per 
sq. in., the allowable stresses may be increased proportionally by 
using the corresponding ratios. 

The allowable unit shearing stresses should be used in deter- 
mining the smallest allowable area of a beam. In beams of large 
proportions carrying considerable load, also in beams subjected to 
dynamic loads, such as those in bridges, the whole length of the beam 
should be provided with web reinforcement even if not required by 
unit stresses. 

Distribution of Diagonal Tension between Concrete and Web 
Reinforcement.—Tests indicate that diagonal tension is resisted 


6 Required anchorage is same as specified under “ Working Stresses, Bars 
with Anchored Ends,” p. 263. 
7 Cylinder strength of concrete at 28 days. 


246 REINFORCED CONCRETE DESIGN 


by the concrete in the web and by the web reinforcement. The 
proportion resisted by each material is not definitely known; there- 
fore, in determining the amount of web reinforcement, part of the 
stresses to be carried by the steel must be assumed as discussed in 
the chapter on Theory of Reinforced Concrete. Two methods of 
distributing the stresses are in common use, namely : 

Method 1.—Conerete takes all stresses when the unit stresses 
are equal to or less than the working unit stress specified for plain 
concrete (same as in method 2). Where the working unit stress 
for plain concrete is exceeded, the concrete is assumed to resist an 
amount of shearing stresses equal to the allowable stresses for plain 
concrete, and the steel is assumed to resist the balance. 


Concentrated 
Loads ~. 
b “I Weight of Beam 








| 
Vo Stirrups | 






















5 = 
Ere ra & 
{ . 
No Stirrups 1S : S 
Required ae NGO ane % 
| by Web sae chal (S 
= Reinforcement © iS ment 5 
| emer ey |S g 
jue Resisted by Concrete ia ae ihe Aesiste Concrete 1S 
Woe ne Ss 
ee | 
a 


Uniformly Distributed Loading Concentrated Loading 


Fic. 79.—First Method. Distribution of Diagonal Tension between Concrete 
and Web Reinforcement. (See p. 246.) 





Method 2.—Concrete takes all stresses when unit stresses are 
equal to or less than the working unit stress specified for plain 
concrete. For sections where this working unit stress for concrete 
is exceeded, the concrete resists one-third and the steel two-thirds of 
the diagonal tension. 

The first method requires somewhat less web reinforcement 
than the second method. The required spacing of stirrups at the 
points of maximum shear is fairly alike in both methods, since here 
the stresses attributed to concrete are about the same in each case. 
The difference is more noticeable at points of small shear, where the 
first method would permit larger spacing of stirrup; but here it is 
apt to be offset by the rule which limits the maximum spacing of 
stirrups. 


FORMULAS FOR DIAGONAL TENSION 247 


Either method may be used with satisfaction. The first method 


is recommended by the authors. 


The 1916 Joint Committee recom- 


mended the second method, while the 1924 Joint Committee recom- 


mends the first method. 


FORMULAS FOR DIAGONAL TENSION 


NOTATION 


V = total external shear, in pounds, at section considered; 
v = total unit shearing stress at section, lb. per sq. in. 


, 


alone, lb. per sq. in.; 


1 
zx 















= 
{ S 
7 No Stirrups EB 
ae Ss Required ~.. > 
! "8/9, 4 5| © 
1 ee ei ee 5 
1 Con> v' 3 
H Resisted by . S 
Web Reinforcement 


Uniformly Distributed Loading 


niles Resisted 


v’ = allowable unit shearing stress (or diagonal tension) on concrete 


Concentrated 
vw boads-. 










Weight of Beam, 
< 





by Concrete Wo Stirrups 
A Required} 


5 mae £ 
Center Line of Span 


raed 
l 3,"2 


by 
‘orcement 


Concentrated Loading 










Resisted 
Wed Rei. 


Fic. 80.—Second Method. _ Distribution of Diagonal Tension between Concrete 


and Web Reinforcement. 


(See p. 246.) 


jf’. = ultimate unit compression stress of cylinder of same concrete 


at 28 days; 


o 
| 


= allowable unit stress in stirrups, lb. per sq. in.; 


A, = cross-sectional area of all legs of a vertical stirrup in square 
inches. (In a U-stirrup this is the sum of the area of the 


two legs) ; 


gd = moment arm, or distance in inches from center of compression 
to center of horizontal reinforcement. (In a T-beam, this 
may be taken, for diagonal tension computation, as distance 
between center of slab and steel; in a rectangular beam, as 
0.87 of the total depth of steel) ; 


~ 
I 


breadth of beam in inches; 


248 REINFORCED CONCRETE DESIGN 


b’ = breadth of web in T-beam in inches. 
s = spacing of stirrups, in inches, at section considered (Figs. 81 
and 82). | 


Unit Stresses.—The unit shearing stresses at any point, where 
the external shear is V, may be found by formulas given below. 
These values should not exceed the allowable stresses given in pre- 
ceding paragraph. 

Shearing Unit Stress in Rectangular Beam, 


Vv 
pid (36) 
Value of jd may be ordinarily taken as id. 
Shéaring Unit Stress in T-beams, 
Vy 
i b’jd a peel ee Pio ck toate ann (37) 


Value of jd may be taken as d — ; but not less than 7d. 


In irregular sections, for b and b’ in above formulas, the smallest 
dimensions should be used. 

Area and Spacing of Web Reinforcement.—Formulas are given 
below for area of cross section and maximum spacing of web members 
at any point in the beam where the external shear is V. Two 
alternate assumptions are given as to the distribution of the diagonal 
tension between the concrete and the web reinforcement. In both 
cases the external shear, V, is the average value in the length of 
beam equal to the spacing of stirrups. 

1. Concrete assumed to resist a definite amount of diagonal ten- 
sion, v’, and the web reinforcement the rest. (See Fig. 79, p. 246.) 

Required Area of Web Reinforcement in a Distance s, 


Sennen 
ine! 


A; 
fs 





s for vertical stirrups. . . . . (88) 





Ales j s for inclined members. . . (89) 


° 


FORMULAS FOR DIAGONAL TENSION 249 


Spacing of Web Reinforcement having a Cross Section, A., 


42518 | ; 
Seis. FOR VENLICAl SUIsTUDS. | sy + yy (40) 
ee ae v’b 
qd 
A sf 
s= 1.435, for inclined members: . . . (41) 
——vb 
jd 


2. Concrete assumed to resist one-third of the diagonal tension 
and the web reinforcement the rest. (See Fig. 80, p. 247.) 


Oe 1a. Stirrup Supports 






Z 






\ s 
~Tensile Bars 





Spacings of Stirrups 
Fic. 81.—Spacing of Stirrup. (See p. 248.) 


Neutral Axis 





Spacing of Bent Bars 
_ Fig. 82.—Spacing of Bent Bars. (See p. 248.) 


Required Area of Web Reinforcement in a Distance, s, 





ee 
1 aad SOT VOEUICa) SULT DS auld ear ee ee 
A, = 047 i, for inclined members... . . . (48) 
Spacing of Web Reinforcement having a Cross Seciion, Az, 
ie a ‘for vertical stirrups. . . . . (44) 
jd 
yee a eas ols for inclined members. . . . (45) 


jd 


250 REINFORCED CONCRETE DESIGN 


The area of reinforcement, As, in these formulas is the sum of 
the areas of all prongs of the stirrup or all bars bent up at one place. 

The above formulas may be used for angles of inclination of the 
bar with the beam axis of from 25 to 50°. For larger angles use 
formulas for vertical stirrups. Tests discussed on p. 42 prove 
that there is no difference in effectiveness between bars bent at 
different angles within this limit. The 1924 Joint Committee 
recommends more elaborate formulas based on the use of actual 
angles. Different: formulas are used for small and large angles. 
This refinement is hardly warranted. 

Limiting Spacing of Stirrups.—In the section of the beam where 
the unit shearing stresses exceed 60 per cent of the maximum allow- 
able unit stress, the spacing of stirrups should not exceed 3d. Else- 
where the spacing should not exceed {d. 





Diagonal tension in this section not provided for 


Fic. 83.—Limiting Spacing of Bent Bars. (See p. 2505) 


Limiting Spacing of Bent-up Bars.—To make the bent bars fully 
effective as diagonal tension reinforcement, the spacing of the bars, 
measured along the axis of the beam, should not be larger than the 
effective depth of the beam. For larger spacing, or where only 
one bar in a beam is bent up, the region of effectiveness of the bent 
bar as diagonal tension reinforcement extends, on both sides, from 
the point of intersection of the bar with the neutral axis for a dis- 
tance along the beam equal to one-half of the depth of the beam. 

The effect of the bent bar in resisting diagonal tension cannot 
exceed the amount of shearing stresses to be resisted within the 
spacing specified above, irrespective of the strength of the bar. 

A bar bent up at a flat angle (around 15° with the horizontal) 
may be considered as effective for a distance equal to twice the dis- 
tance given above. 


FORMULAS FOR DIAGONAL TENSION 251 


Procedure in Designing Web Reinforcement.—Determine the 
maximum total shear, V, and from this the unit shearing stress, v. 
See that v does not exceed the maximum allowable stress. 

Vertical or Inclined Stirrups.—If vertical or inclined stirrups only 
are used, select the diameter and shape of the stirrup so that the 
minimum spacing is not too small (preferably not less than 4 in.), 
and the total number of stirrups in a beam not too large. Remem- 
ber that the maximum spacing of stirrups in the part of beam where 
stirrups are required must not exceed three-quarters of the depth of 
beam. Use the same size of stirrup and the same design for the 
whole length of the beam and, if possible, for all similar beams in 
the entire structure, as a variety of designs may lead to errors and 
confusion. 

Common types of stirrup consist of {-in., 3-in., ~4-in, or 4-in. 
diameter round bars in the shape of a U with the free ends hooked. 

For uniformly loaded beams, the number and the spacing of 
stirrups for different conditions may be taken from the table on p. 900. 

For concentrated loads, a semi-graphical method of determining 
the spacing of stirrups is the most satisfactory. (See p. 252.) 

Vertical Stirrups and Bent-wp Bars—Determine the maximum 
total shear, V, and shearing unit stress, v, as in previous case. 

Determine the number of bars to be bent and the places where 
the bends can be made. (See p. 287.) 

Mark the distances within which the bent bars may He considered 
as effective as web reinforcement. Compute the shear to be resisted 
in these regions and also the available strength of bent bars. Ifthe 
strength of bars is equal to or larger than the shear to be resisted, no 
stirrups are required in these regions. 

Select proper diameter of stirrup, as suggested in previous case. 

Provide stirrups where stresses are not resisted by bent bars. 

If bent-up bars are bent in places where they cannot resist diag- 
onal tension or are bent in one or two places only, their full value 
as web reinforcement may not be available. Bent-up bars may 
be considered as effective web reinforcement for a distance equal 
the depth of the beam. 


252 REINFORCED CONCRETE DESIGN 


SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 


The semi-graphical method suggested below is well adapted for 
practice. The diagrams may be drawn free-hand on computation 
sheets, which, to simplify the work, should consist of cross-section 
paper with squares of about 1_in. diameter. In drawing the diagram, 
each horizontal division may be taken to represent 1 ft. of the span 
and each vertical division 10-lb. of the unit shearing stress. The 
fractions of a foot may be gaged by the eye, so that no additional 
scale is necessary. 

Method of Procedure.—Lay out length of beam to a convenient 
scale. When beam is loaded by concentrated loads, indicate points 
of application of load; for uniform or symmetrical loading, only 
one-half of beam needs to be considered. 

Compute unit shearing stresses (to be considered as measure of 
diagonal tension) at points of maximum shear and, in case of con- 
‘ centrated loads, also at points of concentration. 

Plot the unit shearing stresses as ordinates, to scale, at points of 
maximum shear and also at points of concentration, if any. The 
resulting diagram will be similar to the external shear diagram. 

Mark off the portion of diagonal tension resisted by concrete. 
If concrete is assumed to resist at all sections a definite amount 
of diagonal tension (say 40 lb. per sq. in.), draw a horizontal line 
at a vertical distance from the bottom equal to 40 lb. Then the ° 
upper part of the diagram represents the shear to be provided for 
by stirrups. If concrete is assumed to resist a percentage of total 
shear (say one-third of the shear) mark off, starting from the top, 
the portion resisted by the concrete. The lower part of the diagram 
will represent shear to be provided for by stirrups. . 

Decide on diameter of bar for stirrup and on the number of legs. 

Find spacing of stirrups as follows: From the sketch, scale at the 
support the shear to be resisted by stirrup. From the table on 
p. 899 for proper width of beam, read the spacing corresponding to 
the selected diameter of stirrup and the scaled shear. Mark this 
length on the sketch, starting from edge of support. This is the © 
first division. Next, scale the shear at the end of this first division § 
and proceed in the same manner until all diagonal tension to be 
resisted by web reinforcement 1s taken care of. The stirrups should 
then be placed in the center of each division. 


8 Actually, the shear should be scaled at the center of each division. The 
error of the recommended procedure is slight and on the safe side. 


SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 253 


In practice, the first spacing is repeated several times, then the 
next spacing is duplicated, so that actually only three or four spac- 
ings need to be determined by reference to the table. 

When bent-up bars are used, mark on the sketch their position 
and the distance in which they are effective. Compare the available 
tensile strength of bent bars with the amount of diagonal tension 
to be resisted in the distance in which the bars are effective. Ten- 
sile strength of bar is area multiplied by unit stress in steel, fs, and 
by 1.48. The amount of diagonal tension equals shear scaled at 
point of application of the bar multiplied by width of beam and by 
the distance in which bent bar is effective.. If the tensile strength is 
equal to or greater than the amount of diagonal tension to be resisted, 
no stirrups are then required; otherwise, the diagonal tension not 
taken care of by the bent bars should be provided for by stirrups. 
Elsewhere the same method should be used as when no bent bars are 
used. (For example, see p. 582.) 

Spacing of Stirrups for Uniformly Distributed Loading.—For 
uniformly distributed load, the shear is a maximum at the support 
and zero in the center. The shear diagram to be used in determining 
the spacing of stirrups for one-half of the beam is a triangle, as shown 
in Fig. 253. The required spacing of stirrups obtained in the man- 

ner described above is also shown. 









a 


Resisted by 
Stirrups 


_-- Shear Diagram 


-----------)7-------- — 
Seen ea Se eee 


wat 


: 


Fig. 84.—Spacing of Stirrups for Uniformly Distributed Load. (See p. 253.) 


Spacing of Stirrups.—For concentrated loads, such, for instance, 
as are produced by beams running into a girder, the shear diagram 
used for determining the spacing of stirrups will not always be 
a regular figure, as it depends upon the number of concentrated 
loads and their position. For symmetrical position of concentrated 
loads, the spacing of stirrups is symmetrical at both sides, and one- 


254 REINFORCED CONCRETE DESIGN 


half of the diagram is sufficient. For unsymmetrical position of 
concentrated loads, the spacing will be different at both ends. 

Particular attention is called to the fact that for two beams of — 
equal length, equal concrete dimensions, and same maximum shear 
at the supports, but one of which is uniformly loaded and the other 
loaded with concentrated loads, the spacing of stirrups throughout 
the beams will be different. This is evident from a comparison of 
Figs. 84, 85, and 86, giving spacing for different position of con- 
centrated loads. 


Shear Diagram 






Eq ee Se 
Resisted by 
Stirrups 


‘ 

H 

i] 

1 

' 

‘ 

ss 

( 

' 

> 

ahaa haat 
ress ieee 
ees ere, 
EMG Prize, 


Fic. 85.—Spacing of Stirrups for Load Concentrated in the Center of Beam. 
(See p. 254.) 


| 
_ 


Shear Diagram-~., 


Constant 


PStirrups > quagt Bape ee ee 
~~ Constant Spacing,. 


Resisted by 
Stirrups \ 


— aa ee owe )) Hen nn 
es Vv ae ae ok ee aw es ae 


Fic. 86.—Spacing of Stirrups for Loads Concentrated at Third Points. 
(See p. 254.) 


One Load Concentrated in the Center.—In this case, the shear will 
be uniform for the whole length of the beam (except for the small 
difference caused by varying shear due to dead load of the beam). 
After computing the shearing unit stress and selecting the kind of 


SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 259 — 


web reinforcement, determine the spacing by dividing the tensile 
value of one stirrup by the shearing unit stress to be resisted by the 
stirrup times the width of the beam. ‘The spacing of stirrups will 
then be uniform throughout the beam. 

Loads Concentrated at Third Points.—The shear due to concen- 
trated loads is constant between the supports and the loads. The 
spacing of stirrups there is uniform. No stirrups are theoretically 
necessary in the middle third of the beam. For important beams, 
stirrups are placed at intervals throughout the whole length of the 
beam. 

Concentrated Loads, Irregularly Spaced with Uniform Load.— 
Lay out the span and, at points of concentration, mark off to scale 


2000 16. 15000 1b. 20000 Ib. 5000 1b. 





25000 16. 23000 1b. 


} 
bay OS) 
=~ 
S 





bs Resisted by 
2 Concrete 






174 lb.-- 





© miles 
eee 

a b ’ ' j K 

oe ley spaces @ 10- | vee spaces @ 10~--- @ os | & 

ace amas Resisted by Stirrups Fee 

2 dq e1 Ss 

. Resisted by *5 on 

= de 3 Concrete 14 ? 

ey A 


Fic. 87—Spacing of Stirrups for Concentrated Loads. (See p. 255.) 


the unit shearing stresses (as the measure of diagonal tension). 
Mark off the amount of shearing stresses to be resisted by the con- 
crete. Then, starting at the support, determine and mark off the 
area that can be resisted by one stirrup, and place the first stirrup 
in the center of gravity of that area. Next, mark off the area for 
the second stirrup and place the stirrup and proceed thus until the 
total area of the diagram is provided for. Where the shear does 
not change radically, the same spacing may be repeated several 
times. 

Table 14 on p. 899 will be found useful in determining the 
spacing of stirrups. 


256 REINFORCED CONCRETE DESIGN 


To illustrate the method more clearly, detailed computations are 
given below for determining the number and the spacing of stirrups 
for a beam 20 ft. long, the dimensions of which are 6 = 10 in., 
d = 25 in., jd = 22 in., and the loading as shown in Fig. 87.. 

The reaction is found first. Then the external shears are com- 
puted at points a, 6, ¢, and d. The external shears divided by bjd, 
which in this case is 10 X 22 = 220, give the unit shearing stresses 
at the respective points, as follows: 




















External Shear Shearing Unit Stress 
Points 

Left Right Left Right 
CSA ORI ee 20005) Meee aes 114 
b 24 200 22 000 110 100 
Cc 20 200 5 200 92 23 
d 4 000 —16 000 18 —73 
e —18 800 —21 800 —86 —99 

sf +93. 000. aie 5 ene —104 








The shearing unit stresses are plotted as shown in Fig. 87. 
The length of the beam is laid out first, and the location of the points 
of application of loads marked. The shearing unit stresses are then 
plotted as ordinates at the respective points. After the points are 
connected, an area is obtained which represents the total amount of 
shearing stresses acting on a beam 1 in. wide. The shearing stresses 
at the left are plotted above the line af, and those at the right below 
the line. 

The section in which concrete alone can resist the total diagonal 
tension is found by drawing horizontal lines above and below at a 
distance equal, in the accepted scale, to vo’ = 40 lb. At the left end 
of the beam, this line strikes the outline at point g, and at the right 
end at point di. To the right of point g and to the left of di, all 
diagonal tension is resisted by the concrete and no stirrups are neces- 
sary. The shearing stresses outside of points g and d; must be 
provided for by web reinforcement. 

Assume that steel resists two-thirds of the shearing stresses and 
concrete one-third. Then, to the left of point g and to the right 
of di, one-third of the area is resisted by concrete and two-thirds by 


SEMI-GRAPHICAL METHOD OF SPACING STIRRUPS 257 


web reinforcement. The area resisted by concrete is marked off 
on the diagram. The balance of the area represents the stresses to 
be resisted by steel. 

The total amount of diagonal tension to be resisted by steel at 
the left end of the beam may be found by computing the areas aa;beb 
and 6bicc; and multiplying them by the width of the beam yb =" 10in. 


Areas aajbeb + bbicic = 








2A TOV ie. 2/100 +72\ | 
A : ) x 24 + 3( ) x 60 
= 1792 + 3 440 = 5 232 Ib. 


Multiplying by 6 = 10, we obtain 52 320 Ib. as the total amount 
of diagonal tension to be resisted by the web reinforcement. 

For 3-in. stirrups with two legs, the area of one stirrup is 
A; = 2 X 0.196 = 0.392 sq. in. and the stress resisted by one stirrup 
is Asfs = 0.392 X 16000 = 6 272 lb. 

Divide the total amount of diagonal tension by the stress resisted 
by one stirrup, to get the total number of stirrups; or, 


_ 52320 


N; — 6 272. a 8.3. 


Use 9 stirrups. 

The spacing of stirrups may be found as follows: By trial, start- 
ing at the support, find shear areas in Fig. 87 equal to the resisting 
O74 = 627.2 1b. This 
may be done by scaling the ordinate above the line af, approximately 
in the middle of the first division, and dividing 627.2 Ib. byt. 
Scaling the ordinate at a distance of 4 in. from a, we get 73 Ib. as 
the average unit shearing stress for the first division. The length 
we = 8.6 inches. Lay off this dis- 
tance and place the stirrup in the middle. Next; scale the ordinate 
4 inches from the end of the first division and find the next spacing. 

In this example, as the effect of the uniform load is small, the 
matter is simplified by considering the ordinates between a and b 
and between b and ¢ as constant and finding the spacings for these 
two constant values. Thus, we find the spacing in portion ab to be 
8.5 in., and in portion bc, 10.5 inches. So we may use three spaces 





value of one stirrup divided by b = 10, or 





of the first division, then, is 


258 REINFORCED CONCRETE DESIGN 


at 8 in. and six spaces at 10 in., giving the required number of stir- 
rups. 

The same method may be used in determining the spacing in the 
right end of the beam. The shear here is somewhat smaller than 
at the other end, but it facilitates the erection to adopt the same 
spacing at both ends. Of course, the stirrups must extend to the 
point d. 

Table 14 on p. 899 will be found useful in spacing stirrups. 


DIAGONAL TENSION FOR MOVING LOADS 


For beams carrying moving loads, as bridges and crane runways, 
it is necessary to determine for every section the maximum total 
shear, draw the 
shear diagram and 
space the stirrups 
as suggested above. | 
For heavy moving 
loads, it is advisable 
to make allowance 
for impact. The 
magnitude of the 
impact depends 
upon the character 
of the structure, the 
loading, and the re- 
; BBs Bs lation of the weight 
of the structure to 
the moving load. 
Fig. 88.—Shear Diagram for Uniformly Distributed For railroad bridges 

Moving Load. (See p. 258.) and bridges carry- 

ing electric cars, 

the ordinary impact formulas may be used; for crane runways and 

highway bridges, from 29 to 50 per cent of the live load should be 
added to allow for impact. 

Shear Diagram for Uniformly Distributed Moving Load.— 
Assume moving load, w, per lin. ft.; then the maximum positive 
shear at any section occurs when the load extends from the right - 
support to the section under consideration and the portion between 
the left support and the section is unloaded. (See Fig. 88.) The 


--Dead and Live Load-- 





DIAGONAL TENSION FOR MOVING LOADS 259 


general equation for the maximum shear due to a moving load 


approaching from right to left then isan = saa This is an 
equation of a parabola. | 
wl al! mee ‘a a 
ior. = 0, oe for x = 5, Verte; Foro le VV iO): 


To the shear due to moving loads, the shear due to fixed (dead) 
loads must be added. In Fig. 89, the diagrams for dead and live 
loads are drawn. 
The first line gives 
the diagram for 
dead load only; 
the second line 
for moving load 
plus impact only; 
and third line for 
the sum of the 
dead and live load. 
It will be noted 
that there is con- 
siderable shear at 
the center of the 
beam. 

Stirrups must 
be provided for 
the sum of shears. (6) 

The diagram Fia. 89.—Shear Diagram for Two Moving Loads a Con- 
should be used stant Distance Apart. (See p. 259.) 
only for half of the 
‘span. For the other half, the moving load should be considered 
as the left. The spacing of stirrups must be made the same for 
both ends of the beam. 

Shear Diagram for Two Equal Moving Loads a Constant Distance 
Apart.—This case occurs in a beam carrying cranes and in highway 
bridges (see Vol. III). The maximum shear at any section is obtained 
by placing one load at the section considered. A general equa- 
tion is: ; 

P = concentrated moving load; 
e = constant distance between the loads, P. 








| 


--- Dead and Live Load------ 
—------Live Load------- 


260 REINFORCED CONCRETE DESIGN 


i— (e435) A ees: 
Ve ea or em (46) 


V =P * tors >t—« en 


The variation in shear is a straight line for both equations. It is 
therefore sufficient to determine shear at two points only, namely for 
¢ a Qiatid Col =e: (eee Migs soap 


Maximum shear for x = 0, Vinax = an : | (48) 
and 
Shear for fo madi ne, V= PS. pvt 49) 


With these two values, the shear diagram can be drawn. To 
this must be added the shear due to the dead load. The shear 
diagram having been drawn, the spacing of web reinforcement is 
determined as in previous cases. The diagram is used only for one- 
half of the beam. The stirrups in the other half are made sym- 
metrical. . ; 

For bridge design, the shear may be found as directed in the 
chapter on Bridge Design, Vol. III. The shear diagram may then 
be plotted and the spacing of web reinforcement determined as 
explained before. 


BOND OF STEEL TO CONCRETE 


Importance of Bond Resistance.—It is of the utmost importance 
to design reinforced concrete construction so that the bond stresses 
between the steel and the concrete will not exceed the safe bond 
resistance. The safe bond resistance is equal to the unit stress 
at which the bar begins to slip, divided by the factor of safety. If 
bond stresses exceed safe working values, the factor of safety is low- 
ered and there is danger of separation of concrete and steel and a con- 
sequent failure of the construction from pulling out of the steel. 

Bond stresses do not always receive proper consideration. Many 
designers consider them of secondary importance and often do not 
even compute them. As a result, one often finds designs with 
adequate areas of cross section of longitudinal steel, but with no 
means of bringing the reinforcement into action. 


BOND OF STEEL TO CONCRETE 261 


Bond Stresses Should Always Be Computed.—It is particularly 
important to compute the bond stresses in short beams with heavy 
loads, footings, and retaining walls, where they are likely to be 
excessive unless properly provided for. Bond stresses are often 
excessive also in joist construction, where, because of the narrowness 
of the joist, bars of large diameter are used. 

The importance of proper provision for bond stresses will be 
péalized if it is considered that, ordinarily, the bond is the agency 
which causes cooperation between concrete and steel, and that the 
resistance of steel to tension can be utilized only through the bond 
between concrete and steel. When the bond between steel and con- 
crete is not sufficient, the bar will slip instead of resisting tensile 
stresses. The function of bond in reinforced concrete is the same 
as the function of rivets in built-up structural steel beams. In a 
structural steel beam consisting of web plates and flange angles, the 
angles will not act unless they are properly riveted to the plates. 
When the number of rivets is not large enough to hold the plates 
and angles together, the rivets shear off and the angles become 
separated from the plates. In the same way, in a reinforced con- 
crete construction, when bond stresses are too large, the bars pull 
out instead of acting as tensile reinforcement, and the beam will 
fail even if it is provided with a sufficient amount of tensile rein- 
forcement, because there is no medium which will bind the two 
materials. It would be just as illogical, in designing a reinforced 
concrete beam, to supply tensile reinforcement without proper bond 
as it would be, in steel construction, to supply plates and angles 
for a built-up section without a sufficient number of rivets to make 
the component parts act as a unit. 

Bond is a Function of External Shear.—Consider, in a horizontal 
beam subjected to bending, a portion of beam between two vertical 
sections. Assume that the depth and the amount of steel. between 
_the two sections are constant, but that the bending moment at one 
section is larger than at the other section. The unit stresses in steel, 
being proportional to the bending moment, are larger at the section 
with the larger bending moment. The increase in stresses in steel 
between the two sections must be developed in the bar by the bond 
between the steel and concrete acting in the distance between the 
two sections. From mechanics, the increase in the bending moment 
between two sections is a function of ‘the external shear. The 
increase in stresses in steel will also be dependent upon the magni- 


262 REINFORCED CONCRETE DESIGN 


tude of the external shear. Therefore, finally, the bond stresses 
are a function of the external shear. 
Formulas for Desiga. 
Let V = total external shear at section considered, in lb.; 
y = shearing unit stress in lb. per sq. 1n.; 


u = bond unit stress in lb. per sq. in. of surface area of 
tension steel; 
o = perimeter of one bar in inches; 
So = sum of perimeters of all horizontal tension bars at 


section considered; 
jd = distance between centers of tension and compression 
(approximately $d) ; 
d = effective depth of beam, in inches. 


Then 
Bond Unit Stress, 
mA 
Required Perimeter of Tensile Reinforcement, 
y =< ic 
20m idu’ 2 tel STS oie Pane (51) 


If a portion of tensile reinforcement is bent up, only horizontal 
bars, and those bent-up bars which are not farther up than one- 
third of the effective depth of the beam, should be considered as 
resisting bond at the section under consideration. 

Special attention is called to the fact that in computing bond 
stresses in continuous beams, in the region between the support 
and the point of inflection subjected to the negative bending moment, 
the top bars must be used in the formula. In the balance of the 
beam, the bottom steel must be used. 

Bond Stresses for Small Bars are Smaller than for Large Bars.— 
The ratio of the perimeter of a bar to its sectional area is not constant 
for all sizes of bars. It is larger for small bars than for large bars. 
This means that, for each square inch of sectional area, the sum 
of perimeters is larger for small bars than for large bars. When the 
bars are subjected to equal tension, those having larger perimeter 
will have smaller bond resistance than those having smaller perimeter. 
For this reason, where subjected to equal tensile unit stresses, the 
bond stresses for small bars are smaller than for large bars of the 


BOND OF STEEL TO CONCRETE 263 


same sectional area. This property is often utilized for reducing 
excessive bond stresses. 

The above principle can be clearly seen from the following 
comparison. For instance, 4-in. round bars have an area of cross 
section of 0.196 sq. in., and a perimeter of 1.57 in. The ratio 
1.57 
0.196 
that for 3-in. bars of an aggregate area equal to one unit (say one 
square inch) the total perimeter is equal to 8 units (8 in.). 

For one-inch bar, on the other hand, the area of bar is 0.785 sq. in., 
the perimeter is 3.14 in., and the ratio of perimeter to area is 
3.14 
0.785 
unit has only 4 units of perimeter. From the above figures, it is 
obvious that for equal area of cross section the $-in. bars have twice 
as large a perimeter as the one-inch bars. The tensile stresses are 
governed by the area, and the bond stresses by the perimeter. The 
larger the ratio of the perimeter to the area, the smaller are the bond 
stresses for equal tensile strength. Therefore, for equal tensile 
stresses, the bond stresses for 4-in. bars would be only half as large 
as for one-inch bars. Conversely, the bond resistance of 4-in. bars 
is twice as large as the bond resistance of one-inch bars. 

Working Unit Bond Stresses.—The bond stresses computed by 
Formula (50), p. 262, should not exceed the following values. 

For plain bars, u = 0.04f",, or 80 lb. for 2 000 lb. concrete. 

For deformed bars, u = 0.05f’c, or 100 lb. for 2 000 lb. concrete. 

u — 0.06f’, may be allowed on specially efficient deformed bars. 

Working Stresses, Bars with Anchored Ends.—When the ends 
of the bars, for which bond stresses are computed, are. provided with 
hooks of proper proportions or are extended beyond the point of zero 
bending moment a sufficient length to develop one-half of the strength 
of the bar, the allowable bond stresses may be increased by 50 per 
cent. To comply with this requirement, in simply supported beams 
the bars must be extended the proper length beyond the support. 
In continuous beams, if bond stresses for the bottom reinforcement 
are considered, the bars must be extended beyond the points of 
inflection governed by the positive bending moment, as shown in 
Figs. 98 to 103. If bond stresses for the negative reinforcement 
are considered, the top bars must be extended the proper length 
beyond the points of inflection governed by the negative bending 


between the perimeter and the area is = 8. This means 


= 4. The aggregate area of one-inch bars amounting to one 


264. REINFORCED CONCRETE DESIGN 


moment. The bent bars which are continuous with the bottom 
reinforcement may be considered as fulfilling the above require- 
ment. 

In cantilever footings, to allow the increased bond stresses, all 
bars must be provided with hooks at their ends. 

Problems in Bond.—In design, the following problems may have 
to be solved: 

Problem 1—To determine whether the bond stresses for the 
tentatively selected bars are within working limits. The external 
shear, V, is known. The required area of steel is found from bending 
moment. Bars giving the required area of steel are tentatively 
selected. 

To solve this problem, the perimeter, 0, of one bar is found and 
multiplied by the number of bars. This gives the value 2o. If 
eroups of different sizes are used, the perimeter of each size is found 
and multiplied by the number of bars in the respective group. The 
sum gives the value of Zo. T he bond unit stress is now found from 
Formula (50). If the bond stresses are within working limits, 
the selected bars are satisfactory. IH the bond stresses are larger 
than the allowable working stresses, either of the following methods 
may be used. 

(a) A larger number of bars of smaller diameter, giving the same 
area, may be substituted. This is the most economical method of 
reducing bond stresses. Small bars have a larger perimeter in pro- 
portion to the area than larger bars. This difference may be sut- 
ficient to bring the bond stresses within working limits. For 
instance, six l-in. square bars have an area, As= 6 sq. in. and a 
sum of perimeters, Zo = 24 in. Ten “in, bars have the same area, 
but their perimeters are Lo = 27.5 in. The bond stresses for the 
Zin bars would be about 15 per cent smaller than for 1-in. bars. 

(b) The depth of the beam may be increased sufficiently to 
reduce the bond stresses. The area of steel found for the smaller 
depth, of course, must be retained. This method, since it requires 
more concrete, should be used only when, for any reason, the first 
method is not possible. For example, if the bond unit stress in a 
beam, as found by formula, 1s vu = 120 lb. per sq. in. and the allow- 
able bond unit stress is 100 lb. per sq. in., the necessary increase by 
this method in the depth of the beam would be in the ratio +2% = 1.2, 
keeping same reinforcement to reduce the bond stresses to 100 lb. 
per sq. in. 


ANCHORAGE AND SPLICING OF REINFORCEMENT 265 


(c) The depth of beam and size of bars may be kept the same, 
but the number of bars may be increased. ‘Thus, in the example 
mentioned under (b), the bond stress may be reduced by multiplying 
the number of bars by 1.2. This also is evidently uneconomical. 

(d) A combination of (a) with either (b) or (c) may be used. In 
such a case the bond stresses should be decreased as far as possible 
by the use of bars of small diameter. After that, further reduction 
may be obtained by increasing either the depth or the number of 
bars. 

Problem 2.—To find the bars giving bond stresses within working 
limits. The external shear, V, and the required area of steel are 
known. 3 | 

In solving this problem, the smallest allowable sum of perimeters, 
o, is computed from Formula (51), p. 262. Bars having an aggre- 
gate area of cross section equal to the required area, and having also 
a sum of perimeters equal to or larger than the value, o, obtained 
from the formula, are then selected. The required area and the 
required minimum perimeters being known, the bars can be easily 
selected. | 

‘Problem 3.—To review a design already made, for known load- 
ing, to determine whether the working stresses are exceeded. 

This problem is solved in the same way as Problem 1. 


ANCHORAGE AND SPLICING OF REINFORCEMENT 


Where Anchorage and Splicing is Required.—Anchorage of 
reinforcement is required when the reinforcement at the face of the 
support is subjected to stresses which must be transmitted from the 
bars to the support. The stresses from the bars may be transferred 
either to the concrete of the support or to another set of bars within 
the support. As an example, we may consider the bars at the sup- 
port in continuous and restrained beams and in cantilevers, because 
there the maximum tensile stresses in steel, produced by the bending 
moments, exist at the face of the support and must be transmitted 
to the support. Another instance of anchorage is seen in the splicing 
of two bars at a point where the bar to be spliced is subjected to 
considerable tensile stress, and it is desirable to transmit the stress 
to the other bar. 

Means Used for Anchoring Bars.—The simplest means of anchor- 
ing a bar is by extending it from the beam into the concrete of the 


266 REINFORCED CONCRETE DESIGN 


support, for such a distance that the stresses from the bar are trans- 
ferred gradually, by bond, to the concrete of the support. (See 
Fig. 90.) | 

Another method of anchoring a bar is by providing it at the end 
with a hook of sufficient dimensions. The hook is placed within the 
support. 

The best method of anchorage is to extend the bar some distance 
into the support and also to provide it with a hook at the end. (See 
Fig. 90.) 


-Full Strength of bar 
at this Section at Crack may open 









Correct Anchorage as Wrong Anchorage 


é Bars should overlap 
Nor be continuous 


PE i PEE A 






| yi Grack may open 
ED BLO natn a vee semiais ERE S At 
CU tsi an sake oe Ti ecu dene fetus ost 


POLS Me We eee es 
Ver oe ~ eaeeie 
SSeS ie 


Correct Anchorage Wrong Anchorage 


Fra. 90.—Method of Anchoring Bars. (See p. 266.) 


Method of Splicing Bars.—The stress from one bar may be 
transferred to another bar by splicing. The splice may be formed 
by lapping the bars for a length equal to the length of imbedment, 11, 
required for anchorage. A splice may also consist of a straight lap 
with hooks at both ends of bars. The length of the lap may then 
be shorter than for straight ends. 

Sometimes it may be necessary to splice two bars by means of a 
third bar as shown in Fig. 91. The third bar must lap each bar a 
length sufficient to develop the strength of the bar. The length of 
the short bar is then equal to twice the length of imbedment, [1, 
required for anchorage. 

When several bars in a beam or slab are spliced, the splices 


ANCHORAGE AND SPLICING OF REINFORCEMENT 267 


should be arranged so that not more than one bar is spliced at any 
one place. 

Splices at points of maximum stress, for example, in the middle of 
beams, should preferably be avoided. This does not mean, however, 
that they may not be permissible with proper staggering. 

When tensile bars are spliced by lapping, the length of the lap- 
should be based on the maximum tensile stress in steel, even if the 
stresses at the points of the splice are smaller. 


é fs -- Length of Splice tic | @ ) 
o 2SGe S50 TT | 


Straight Splice Splice with Hooked Ends - 














. Length of Length of | 
 fadeSnlice's i ay a Splice" 
Splicing two Bars by a Third Bar 
Fig. 91.—Method of Splicing Bars. (See p. 266.) 


Problems to be Solved in Connection with Anchorage.—The 
following problems may have to be solved in connection with anchor- 
age of the bar: , 

1. Diameter of bar and stress are given, and it is necessary 
to deternmrine the means of anchoring the bar. 

2. Stress, diameter of bar, and length of imbedment are given, 
and it is necessary to determine the bond unit stress. 

3. Length of imbedment, diameter of bar, and bond unit stress 
are given, and it is necessary to determine the safe tensile stress 
to which a bar may be subjected without exceeding the allowable 
bond stress. 

4. The length of the lap or splice for a bar may have to be deter- 
mined. 

These problems are solved by use of formulas and _ principles 
given below. 

Formulas for Determining Anchorage by Straight Imbedment.— 

Let f; = tensile or compressive stress per square inch in bar; 

7 = diameter of bar in inches; 
u = bond in pounds per square inch of surface; 
1; = necessary length of imbedment of bar in inches. 


268 REINFORCED CONCRETE DESIGN 





Then 
Length of Imbedment of Square and Round Bars, 
borspl Hevea im 
iis ay eye . a 
Bond Unit Stress, 
Shoes 
w= at 0 | ph panera ee 


Allowable Tensile Stress in Bar, 


fa = Ale. ee 


Conditions in Beams at Support.—At an interior column ‘in con- 
tinuous beams, the tensile bars at the support are extended from one 
span into the adjoining span. The extended portion acts not only 
as anchorage for the stresses developed in the first span, but also as 
tensile reinforcement in the adjoining span. The length of the 
extended portion of the bar is governed not by anchorage require- 
ments, but by the moment requirements in the adjoining span. 
Good practice in such cases requires that the bars be extended 12 in. 
beyond the point of inflection of the adjoining span. 

If a beam extends beyond the column, forming a cantilever, the 
same conditions exist. 

In the end spans of continuous beams, and in restrained beams 
where it is not possible to extend the bars into the adjoining span, 
the length of imbedment is determined solely by the requirement for 
anchorage. It is often impossible to get sufficient length of imbed- 
ment, and the bar is provided with a hook. The support in which 
the bar is anchored must be strong enough to withstand the tensile 
stress transferred by the anchored bar. If the support is a column, 
provision may be required for tensile stress in the column, due to 
bending. The tensile reinforcement for the beam and column 
must be arranged so as to prevent cleavage between the two 
members. 

Allowable Bond Unit Stresses for Anchorage and Splices.— The 
allowable bond unit stresses to be used in computing the anchor- 
age and splices are the same as recommended on p. 263 for beams. 

Anchorage of Compression Reinforcement.—In continuous beams 
at the support, bottom reinforcement. is subjected to compressive 
stresses. If the bars are intended to serve as compressive reinforce- 


ANCHORAGE AND SPLICING OF REINFORCEMENT 269 


ment, provision should be made for transferring the compressive 
stresses in the bars at the face of the support into the column. For 
this purpose, the bars are extended a sufficient length into the sup- 
port to gradually transfer the compressive stresses, by bond, from 
the bar to the concrete of the column. These compressive stresses 
act at right angles to the main column stresses and therefore they do 
not affect its capacity as a column. 

The necessary length of imbedment, 11, may be found by Formula 
(52). The stress, f,, to be used in computing the length, is the 
compressive stress in the bar, which is usually much smaller than 
the allowable tensile stress in the bar. 

Sometimes, to increase the amount of compression reinforce- 
ment at the support of a beam, the bottom bars from the adjoining 
beam are extended into it. The extension then serves both as 
anchorage in one span and as compressive reinforcement in the other. 
The length in the adjoining span will be governed by the moment 
requirements. 

Anchorage of Column Reinforcement.—Compression reinforce- 
ment in columns carries considerable stress. Since the bars ordi- 
narily are only one story high, it is necessary to transmit the stresses 
from the bars in the column above to the bars in the column below. 
The requirements are treated in the column section. 

Use of Hook as Anchorage for Reinforcement.— When it is not 
possible to extend the bar sufficiently into the support to develop 
its strength by straight imbedment, the anchorage is accomplished 
by a hook. The effectiveness of the hook depends upon the strength 
of the concrete, as the hook transfers the stress from the bars to the 
concrete by bearing on the concrete. The dimensions of the hook 
must be such that the bearing stresses do not exceed allowable 
working values. Where considerable stress is transmitted by a 
hook, it is advisable to place a cross bar of proper length inside of 
the hook, to distribute the bearing stresses to a large area of con- 
crete and to prevent splitting of the concrete. It is not essential 
that the hook should bear perfectly on the cross bar. Even if there 
is no contact between the hook and the bar, the cross bar will pre- 
vent splitting of the concrete. 

The conditions are still more improved if the cross bar, instead of 
being straight, is inclined at an angle to the direction of the bar to be 
anchored, as shown in Fig. 92, p.270. Then part of the pull may be 
taken by the cross bar in tension and transferred gradually to the 


270 REINFORCED CONCRETE DESIGN 


concrete. Another instance of effective anchorage is found in the 
Smulski Flat Slab System at the column head, where the pull from 
radials is transferred partially by bearing on the concrete and par- 
tially by tension in the center ring. 

When the reinforcement in another beam is at right angles to the 
reinforcement requiring anchorage, it may be considered as assist- 
ing in the distribution of the bearing stresses on the concrete. 

The best anchorage is obtained by a combination of straight 
imbedment equal to 15 diameters of the bar and a semicircular hook, 
the inside diameter of which is at least four times the diameter of 
the bar. (See Fig. 92, p. 270.) With such a design, the elastic 
limit of the steel can be reached without causing excessive secondary 


pecan cal eae 15 Bar Diameters 
a ae _--~;--10 Bar. Diameters 


y- Section of Maximum 
ta . 
—“~4 Bar Diameters 


a ension 
Elevation 









Elevation 


; Cross Bar } 
Plan Plan 


Fic. 92.—Hook Recommended for Anchorage. (See p. 269.) 


stresses in the concrete. When cross bars are used, the straight 
imbedment may be reduced to 10 diameters of the bar. 


BEARING STRESSES 


Let P = superimposed load in pounds; 
A = total area of pedestal or footing (subjected to limitation 
according to depth) ; 
A, = bearing area, in same units as area A; 
fy = allowable bearing unit stress, lb. per sq. im? 
fri = allowable bearmg unit stress on wall, lb. per sq. in.; 
f’. = ultimate compressive strength of cylinder, Ib. per sq. in. 


Explanation of Bearing Siresses.—Bearing stresses are produced 
by a supported member, resting upon a supporting member, at the 
surface of contact of the two members. Bearing stresses are pro- 
duced under the bearing plate or base of a steel beam or column 


BEARING STRESSES 271 


resting on concrete. A concrete column resting upon a_ pedestal 
also produces bearing stresses in the concrete of the pedestal, the 
magnitude of which may be computed by dividing the column 
load by the gross area of the column. When the stress from vertical 
column reinforcement is transferred to the pedestal or footing by a 
plate, the stresses produced under the plate may be called bearing 
stresses. 

The magnitude of bearing stresses equals the total superimposed 
load P divided by the area of contact, 41. Therefore 


P 
fo = Ge (55) 

Allowable Unit Stresses.—The allowable bearing unit stress 
depends not only upon the strength of the material composing the 
supporting member, but also upon the relation between the size of 
the bearing area and the total area of the top of the supporting 
member. If the bearing area equals the area of the top of the 
supporting member, the allowable working unit stress over the 
bearing area must not exceed 0.25f’.. For 2000 lb. concrete the 
allowable bearing unit stress is 500 Ib. per sq. in. 

If the area of the pedestal is larger than the bearing area and 
projects on all sides beyond the bearing area, the allowable unit 
stresses may be found from formula below. 

Allowable bearing unit stress, when area A extends on four sides 


outside area A1. 
/ = A 
fo= 0.25f". Aj ° e ° ° ° ° ry (56) . 


but not more than 0.7f’c. 

The thickness of the footing, pedestal or pedestal and the footing 
must not be less than six-tenths of the maximum projection of the 
area A, beyond the bearing area. 

If a column rests directly on a shallow footing, the total area of 
the footing can not be used in the formula. Instead an imaginary 
area should be used, the sides of which equal the sides of the bearing 
area plus 12 times the depth of the footing. This imaginary area 
must not in any place come outside of the actual area of the footing. 

Formula 56 applies if the bearing area is surrounded on all sides 
by concrete of the pedestal or footing. If the bearing area on one 
or two sides coincides with the sides of the supporting area, as in 


272 REINFORCED CONCRETE DESIGN 


case of a bearing plate resting on a wall, only one-third of the mcrease 
‘n allowable bearing stresses over 0.25 f’: as given by Formula (57) 
may be used. The formula changes to 

Allowable bearing unit stress when side of bearing area A1 coincides 
with side of supporting area A, 


5 
yet 9 0.25/'.(0.67+] 4). gia ie 
but not more than 0.5f’c. 


PROTECTIVE COVERING FOR REINFORCEMENT 


Purpose.—The concrete outside of the reinforcement of concrete 
members serves several purposes. First, it bonds the concrete and 
the steel. In computing bond resistance, the whole perimeter 
of the bar is considered as resisting bond stresses. This is only 
possible where there is sufficient concrete around and below the bar. 
Second, it protects the reinforcement from fire. Third, it protects 
‘the reinforcement from weather and moisture and therefore from 
rusting. Since the amount of concrete that must be placed below 
the steel for protection is always sufficient for bond, only the pro- 
tective purposes will be treated. 

Fire Protection—The thickness of concrete necessary for pro- 
tection of reinforcement against fire depends upon the quality of the 
aggregates, and upon the severity of the possible fire to which the 
structure may be subjected. Aggregates having low coefficient of 
expansion, such as limestone and traprock, are desirable for fire 
protection. Gravel and granite should not be used as an aggregate 
where fire hazard is large, because the quartz, of which they largely 
consist, has a high coefficient of expansion and, under high heat, will 
expand and crack the concrete. 

Limestone and Traprock.—With aggregates such as limestone or 
traprock, the protective concrete, measured from the surface of the 
concrete.to the surface of the lowest bar, should be, for large fire 
hazard: 

Slabs and walls, 1 in. 

Beains, girders and columns, 2 in. 

Where the fire hazard is small, the protective covering may be as 
follows: 

Slabs and walls, 2 in. 

Beams, girders, and columns, 13 in. 


CLEAR DISTANCE BETWEEN PARALLEL BARS IN A BEAM 273 


Gravel and Granite.—When gravel or granite aggregates are used 
in buildings with small fire hazard, the larger protective covering 
recommended for the traprock should be used. 

‘When it is absolutely necessary to use these aggregates in a build- 
ing exposed to severe fire hazard, the protective cover should be 
1 in. larger than recommended for traprock and should be reinforced 
with metal lath placed 1 in. from the surface. The openings in 
metal lath should not exceed 3 in. 

Moisture Protection.—Reinforcement in footings should be pro- 
tected with a minimum covering of 3 in. Where concrete is exposed 
to weather, the concrete covering for the reinforcement should be 2 
in. thick. In all cases, the thickness of the protective covering is 
measured from the face exposed to moisture to the nearest surface 
of the bar. 


CLEAR DISTANCE BETWEEN PARALLEL BARS IN A BEAM 


Parallel bars in a beam must be placed so far apart that the 
amount of concrete around each bar is sufficient to provide proper 
bond between steel and concrete. Proper clear spacing between 
bars, depending upon the maximum diameter of coarse aggregates, 
is also required to allow the concrete to flow below the bars. The 
width of the stem ‘of a T-beam is often governed by the distance 
required for spacing of bars. | 

Spacing Recommended by the Authors.—The minimum clear 
distance between parallel bars should not be less than 1 in., or less 
than 14 times the diameter of the larger round bar and 2 times the 
side of a square bar. 

The minimum clear distance of the bar from the side of the beam 
should not be less than required for protective covering, nor less than 
14 times the diameter of the bar. As a further requirement, the 
minimum spacing of the bars should be larger than the maximum 
diameter of the coarse aggregate. If aggregates cannot pass freely 
between bars, there is danger that arching of the aggregates, will 
produce voids in the concrete at the level of reinforcement. Such 
voids are dangerous because they reduce bond and also reduce the 
effectiveness of the protective covering. 

The table on p. 274 gives the required width of the beam for dif- 
ferent number and size of bars. 


274 REINFORCED CONCRETE DESIGN 
Width of Beam Required by Different Number of Round and Square Bars 


Assumptions: Fireproofing 13 In. to Face of Bar. Clear Space between Bars 
Equal to 1.5 Diameters of Round Bars and 2 times Side of Square Bars. 


ne aS 


Width of Beam in Inches 





























Diam. 
Bars, Number of Bars in One Layer 
In. 
1 2 3 4 5 6 7 8 9 10 
a eee 
Round Bars 
I 
7 | BPH ia pee eet 6.01 7.2 | 8.5 | 9.7 | Si:O8 ee roe 
3 3.02 | toe 6.7 | 8.3 | 9.9 | 11.4°1-18,0 ) Oe SiGe etree 
3 Seen po 7.5.12 9,.4 | 11,2218: 14) 15209 2620) eae 
4 3:9) ie orl 8.2 110.4 | 12.6 | 14.8 | 17.0 | 19.2) 284 2576 
1 470 | 4625 9.0 | 11.5 | 14.0 | 16.5 | 19.0 | 21.5 | 24.0 | 26.5 
1} Aoi Goo 9.7 | 12.6 |.15.4 | 18.2 | 2130 1 2358 e20rea ees 
1} 4.2 | 7.4°| 10.5 |13:6°) 16-7 | 19199792300 (226s 2o ee 
13 4.5 | 8.2 | 12.0 | 15.7 | 19.5 | 23.2 | 27.0 | 30.7 | 84.5) 38:2 
ao ee a ee 
Square Bars 
eee 
3 Bb aioed 6.5 | 8.0:1° 9.5 | 11. ON R22 S50 SOR) Tose 
5 3 Os) Pose 7.41 9,2 111.1 | 18.0 171479 yl Gay Ste eee 
2 ae eieieO10 8.2 | 10:5 | 12.7 | 15.0 | 17.29) T9050 aie aieeee 
rf 3.9 | 6.5 9.1 112.7 | 14.4 1 17.0 | 190622025 eae ee 
1 4.0:| 7.0 | 10.0.| 13.0 | 16.0 | 19.0 22.05 257 0S eet ete 
1} A.1 | 7.5 | 10.9 | 14:0 | -17.6°| 2150) 245s S27 aera 
4 4.2 | 8.0 | 11.7 | 15.5 }-19:2 | 23.0 | 2657) ))300 "eee eee 
13 4.5 | 9.0 118.5 | 18.0 | 22.5 | 27.0 NSLeSpiaeae0 es eeltoee 


Vertical Spacing of Layers of Reinforcement.—Whenever pos- 
sible, bars should be placed in one layer. If more than one layer is 
necessary, the vertical spacing of the layers in the clear should be not 
less than 1 in., or less than the diameter of the largest bar. 

If the design requires the extreme number of three layers of bars, 
the bars in the lowest layer should be properly anchored at the ends 


, BENDING MOMENTS FOR USE IN DESIGN 275 


by hooking or extending into the supporting column. At least one- 
half of the bars, by area, should be bent up and anchored in the com- 
pressive zone, not only to serve as diagonal tension reinforcement, 
but also to develop the tensile stresses by anchorage in the com- 
pressive zone and thereby to insure proper action of the reinforce- 
ment. 

Keeping Bars in Place.— When the spacing of bars is the minimum 
allowable, it is important that the bars should not be even slightly 
misplaced during construction. For this purpose, some positive 
means of separating the bars, such as spacers, should be used. This 
is particularly important in narrow beams. 

Layers of bars must be separated vertically by mechanical means. 
One of the accepted means consists of placing across the reinforce- 
ment short pieces of bars of proper diameter, spaced not more than 
3 ft. on centers. : 

Bars should be kept the proper distance above the forms. Cer- 
tain mechanical spacers keep the bars a proper distance above the 
form and a proper distance apart. Their use is recommended. 


BENDING MOMENTS FOR USE IN DESIGN OF REINFORCED 
CONCRETE BEAMS 


The bending moments for use in design of reinforced concrete 
beams depend not only upon the length and the loading of the span 
under consideration, but also upon whether the beam is a single span 
or is built continuous with a number of other spans. 

In a single span, the question to be considered is whether the 
beam is simply supported, or restrained at the supports. The qual- 
ity of the restraint is also to be considered. The beam may be fully 
restrained, 1.e., fixed, at the support; it may be only partly restrained 
by a concrete column; or it may be slightly restrained by masonry. 

In a continuous beam consisting of a number of spans, it is impor- 
tant to determine whether the spans are equal or not; whether the 
beams are connected with intermediate columns; whether the 
loading of all spans is the same; and finally, whether the ends of 
the beam are freely supported, restrained, or partly restrained. 

Quality of Restraint at Ends.—When an end of a beam rests 
on the support and is free to turn, it should be considered as freely 
supported, and no provision is necessary for negative bending 
moment. 


276 REINFORCED CONCRETE DESIGN 


When an end of a beam runs into a masonry wall and is encased 
in it, especially when additional masonry is placed upon it, the end 
is not free to move as in the previous case. The masonry offers 
restraint to the beam, which should be considered as slightly 
restrained. The amount of such restraint is uncertain and unre- 
liable, however, and it cannot be counted upon as reducing the posi- 
tive bending moment. To prevent cracks at the support, some pro- 
vision should be made for a possible negative bending moment by 
using negative bending moment reinforcement. The amount of 
reinforcement should be governed by the actual conditions, but 
should not be smaller than required by formula on p. 278. The 
bars must be anchored at the end, and at least one-third of the 
reinforcement should extend to a point distant 0.2 / from the edge 
of the support. 

When the end of a beam runs into a flexible support, such as a 
column, and is rigidly connected with it, the end should be con- 
sidered as restrained. The magnitude of this restraint depends 
upon the relative ratio of stiffness of the beam and the column, as 
fully explained in Volume III. Such restraint is reliable and may 
be counted upon in determining bending moments. 

When the end of a beam runs into a support of much larger stiff- 
ness and is rigidly connected with it, it may be considered as fixed 
at that end. A beam may be considered as connected with a column, 
wall, or other support, when the connection is capable of resisting 
the bending moment produced at the support. 

Difference between Continuous Beams and Building Frames.— 
When a continuous beam simply rests upon the intermediate sup- 
ports—as, for instance, a continuous bridge simply resting upon the 
piers and not monolithically connected with them,—or when the 
beam is supported by small intersecting girders, full bending moment 
from one span is transferred directly to the adjoining span. Such a 
continuous beam is sensitive to unequal loading. ~The positive 
bending in such a case should be based upon the possibility of unequal 
loading of the adjoining spans, and the Formulas (63) to (68), under 
the heading “Continuous Beams of Equal Spans,”’ should be used. 
Continuous beams are shown in Fig. 93, p. 277. 

When a continuous beam is connected with intermediate columns, 
as shown in Fig. 94, the bending moment from one span is not 
transferred directly in full to the adjoining span, but a part of this 
bending moment is transferred to the columns, so that the effect of 


BENDING MOMENTS FOR USE IN DESIGN 277 


the loading in one span upon the adjoining span is much smaller 
than for the continuous beams just described. Smaller provision 
is necessary for unequal loading and, for the positive bending moment, 
the Formulas (69) to (74) under ‘‘ Continuous Beams of Equal 
Spans Running into Columns,” may be used. 


Continuous Beam - -, 





Fic. ¢3.—Continuous Beams. (See p. 276.) 


The subject is thoroughly treated in Volume III, to which refer- 
ence should be made for full explanation of the subject. 

Span Length.—The span length for a freely supported beam or 
slab should be taken equal to the clear distance between the sup- 
ports, plus the depth of the beam or slab. 

The span length for a continuous or restrained beam should be 
taken as the clear distance between the faces of the support. 





Fic. 94.—Building Frame. (See p. 276.) 


Use of Formulas.—The formulas given below may be used for 
equal or nearly equal spans and for uniform loading. Where spans 
are not equal or the loads on different spans are not uniform, formulas 
given in Volume III should be used. 


NOTATION 


Let M = bending moment; 
w = uniformly distributed load, lb. per lin. ft.; 
1 = span length of beam in ft., as defined above. 


278 REINFORCED CONCRETE DESIGN 


FORMULAS FOR SINGLE SpAN BEAMS 


Ends Freely Supported.—Beam free to turn at support. 


Maximum Positive Moment at the Center, 
wi? ; 
mire ft.-lb. = 1.50l? inelbs ee 


No Negative Reinforcement is Required, 
Beam Slightly Restrained at Ends.—Ends not free to turn. 


Mazimum Positive Moment, 
wilt oe 
M = ce {tIb. ='1/5ul? ih -be 
Negative Bending Moment at End Supports, 
'F F 
not less than M = oe ft.-lb.-= 0.60wl? in.-lb.3 0 2) OU) 


Beam Fixed at Ends. 
Maximum Positive Moment at Center, 


wl? ; : 
M = 17 ft.-lb. = wl? in.-lbs. of ak ee 
Negative Moment at Supports, 
wl? 
= a ao, ft.lb. = — wi? in-lb. |. eee 


CONTINUOUS BEAMS OF EQUAL SPANS 


These formulas are adapted to such constructions as ordinary 
slabs in beams and slab floors, beams supported by girders, and roof 
construction, as shown in Fig. 93, p. 277. 

The ends may be considered as slightly restrained. 

Beams and slabs continuous for two spans only. 


Mazimum Positive Moment near Center, 
wl? , 
Me 0 ft.-lb. = 1.2wi2anelbp a eee 
Negative Moment over Interior Support, 


wl? : 
1 = mp ft.-lb. = 1/5wl? in loan ee 


CONTINUOUS BEAMS OF EQUAL SPANS 


Negative Moment at End Supports, 


2 
M = not less than Oe {t.-lb. = 0.60wi? in.-lb. 


Beams and slabs continuous for more than two spans. 


(a) Interior Spans 


Maximum Positive Moment near Center, 
Negative Moment at Support, 


pean" rie eee 
(ie 19 ft.-lb. = wi? in.-lb. 


(b) End Spans. 


279 


(65) 


(66) 


Maximum Positive Moment near Centers of End Spans, 


Negative Moment at First Interior Support, 


2 
dee ee pele ale ink inelb: 


Negative Moment at End Supports, 


(67) 


2 
M = not less than ft-lb. = 0.60wil?in.-lb. (68) 


(Building Frames) 


CONTINUOUS BEAMS OF EQUAL SPANS, RUNNING INTO COLUMNS 


Beams and slabs of equal spans built to act integrally with 


(a) Interior Spans. 


Negative Moment at Interior Supports except the Furst, 


ae pacts 
Mex 12 ft.-lb. = wi? in.-lb. 


intermediate and end columns, walls, or other restraining supports, 
and assumed to carry uniformly distributed loads, as shown in 
Fig. 94, p. 277, should be designed for the following moments at 
critical sections: 


(69) 


Maximum Positive Moment near Centers of Interior Span, 


wil? ; 
ve 16 ft.-lb. = 0.75wl? in.-lb. 


(70) 


280 REINFORCED CONCRETE DESIGN 
(b) End spans of continuous beams for which the ratio of stiffness, 
9 is less than the sum of the ratios of stiffness for the exterior 
columns, above and below a rigidly connected beam. 


Maximum Positive Moment near Center of Span and 
Negative Moment at First Interior Supports. 


wl? ; 
M 12 ft.-lb. = wi? in.-lb.3, Wiees ee 
Negative Moment at Exterior Supports, 
wil? 
Mi= 17 ft.-Ibs= wi? inelb. eee ee 


(c) End spans of continuous beams for which the ratio of stiffness, 
- is equal to or greater than the sum of the ratios of stiffness, 
a for the exterior column above and below a rigidly con- 
h 
nected beam. : 
Maximum Positive Moment near Center of Span and 
Negative Moment at First Interior Support, 


wl? me | 
Mo= Tz ft-lb. =] lw? m-th eee 
Negative Moment at Exterior Support, 
2 
M =“ ft-lb. = 0.7502 inlb, - . 12 (7) 


The formulas under (c) are adapted to ordinary floor construction 
where the beams run into column and concrete walls. 


Continuous Beams or UNEQUAL SPANS OR WITH NON-UNIFORM 
LOADING 


Continuous beams with unequal spans or with non-uniformly 
distributed loading should be designed for actual bending moments 
as directed under proper headings in Volume III. With unequal 
spans, provision should be made for bending moment in the columns. 


9'Table on p. 945 gives formulas for moments of inertia, J for various cross 
sections. 


DETAILS OF CONTINUOUS BEAMS . 2381 


Bending moments for concentrated loads are also fully treated 
in Volume III. 


EFFECT OF VARYING MOMENT OF INERTIA UPON THE BENDING 
MOMENT 


As the bending moment in a continuous beam depends upon its 
moment of inertia, it will vary with the variation in the moment of 
inertia. The moment of inertia is seldom constant for the whole 
length of a continuous reinforced concrete beam. - This is especially 
true of T-beams and of beams provided with haunches. 

However, a thorough study of different conditions, by the authors, 
shows that the greatest variation in moment of inertia in a beam 
that is possible in ordinary design causes a variation in bending 
moment of less than 10 per cent. Under most conditions, the varia- 
tion is even smaller. Consequently, a continuous beam may be 
designed safely with the bending moments recommended on p. 279. 
The effect of varying moment of inertia is fully treated in Volume III. 


DETAILS OF CONTINUOUS BEAMS 


Continuous Beam at the Supports.—In the past, too little atten- 
tion has been paid to the details of reinforced concrete beams at the 
supports, with the result that a number of reinforced concrete 
structures have been built with beams and girders containing insuf- 
ficient steel at the top of the beam over the supports to take the pull 
caused by the negative bending moment, and insufficient area of 
conerete to take the compression.!° Not only do these beams fail 
to have the required factor of safety, but frequently even the working 
loads cause cracks that are always unsightly and sometimes dan- 
gerous. Moreover, moisture is likely to penetrate the open cracks 
and rust the steel. 

Just as much care, therefore, is necessary in designing the rein- 
forced concrete beam at the supports as in the middle of the span. 
Not only the tensile stresses in the steel, but also the compressive 
stresses in the concrete must be computed and properly provided 
for. 

In T-beams particularly, there is a likelihood of the concrete being 
overstressed at the supports. At the center of the span, the portion 
of the slab forming the flange is available for resisting compression. 

10 See p. 282. 


282 REINFORCED CONCRETE DESIGN 


At the supports, however, where the bending moment is negative, 
the flange is in the tensile part of the beam and the compressive area 
consists merely of the part of the stem below the neutral axis. Usu- 
ally, the compression area is not sufficient and it is necessary to 
increase it by using compression reinforcement. 

Allowable Working Stresses at the Support.—As is evident from 
the curves of bending moments, the negative bending moment 
decreases very rapidly, so that the extreme fibers are under maximum 
stress only in a very short length of the beam. It is therefore per- 
missible, at the supports, to use a higher compressive working stress 
in the extreme fiber than is used in the center of a beam where the 
rate of decrease in bending moment is much smaller and a larger 
part of the beam is under high stresses. The 1924 J oint Committee 
recommends an allowable working stress in concrete at the support 
of 900 lb. per sq. in. for 2 000 lb. concrete, when the working stress 
in the center of the beam is 800 lb. per sq. in. The authors indorse 
this recommendation. If other unit stress is used in the center, 
it may be increased at support by 15 per cent. 

For a stress in the center of 650 Ib. per sq. in. as specified by most 
Building Codes, a stress of 750 lb. per sq. in. is allowable at the 
support. 

Even with this allowance, it is often impossible, without special 
provisions, to keep the stresses within working limits. 

Compression Steel at the Support.—Frequently, additional com- 
pressive area is provided by reinforcement in the compressive part 
of the beam so that the beam must be considered and computed as 
reinforced at the top and bottom. The effectiveness of steel in com- 
pression has sometimes been questioned, but the results of tests with 
reinforced concrete columns, as well as with beams with steel in top 
and bottom (see p. 49), have proved conclusively that compression 
steel not only takes its share of the stress, but also stiffens the beam. 

First Method. (See Fig. 95).—In a reinforced concrete beam, 
part of the bottom steel is bent up, and the rest, usually composing 
one-half of the steel area, is carried horizontally to the supports. 
This steel may be utilized as compression reinforcement. As the 
bars get their maximum compression at the edge of the support, to 
be effective as compression reinforcement they must be extended into 
the support, a sufficient length beyond the edge to develop the stress 
by bond. (See p. 268.) A shorter length of imbedment for bond is 
required for compression than for tension, because the compression 


DETAILS OF CONTINUOUS BEAMS 283 


stresses are smaller. This is the simplest method of providing com- 

pression reinforcement and should be used where possible. It is 

illustrated in Fig. 95, p. 283. The ratio of compression reinforce- 

ment, p’, equals about one-half of the ratio of the tensile reinforce- 
fe 


ment, or Beh, 
Pi 





-+:2| _----Length of imbedment for compression----~.._ «<— 


Fig. 95.—Compression Reinforcement at Support. First Method. 
(See p. 283.) 


Second Method. (See Fig. 96.)—When the amount of compres- 
sion reinforcement provided, as explained above, by the straight bars 
is not sufficient, additional reinforeement may be obtained by 
extending into the span under consideration the straight bars from 
the adjoining span. The length of the extended bar, measured from 
the edge of the support, will depend upon the bending moment, but 


NN, SS So —_—= = oe oes ee oe 





“\Z_____---Compression Steel from-- 
i Adjoining Span pie: 


Fig. 96.—Compression Reinforcement. Second Method. (See p. 283.) 


it must not be smaller than required to develop by bond, the com- 
pression stresses at the edge of the support. In turn, the straight 
bars from the span under consideration are extended into the adjoin- 
ing span so that the compression reinforcement at both sides of the 
support consists of the sum of the horizontal bars in both spans. 
Such an arrangement is shown in Fig. 96, p. 283. The ratio of the 
compression reinforcement at the support, p’, is thus about equal to 


, 
the ratio of tension reinforcement, pi. Thus a = |. 
1 


Since each horizontal bar is subjected to compression stresses 
‘in the original span and also to equal compression stresses in the 
adjoining span, a question may be raised about the amount of com- 


284 REINFORCED CONCRETE DESIGN 


pression stresses carried by the bar. The answer can be had by 
referring to Fig. 97. From this, it is obvious that the compression 
stresses in one span act in the opposite direction to the compression 
stresses in the adjoining span. Thus, the compression stresses in the 
bar from the two spans are not added but balanced. The compres- 
sion stress in the bar in one span acts as a reaction to the compression 
stress in the other span, and the stress 1s the same as though only one 
span were acting. 

Third Method—Sometimes the amount of compression reinforce- 
ment obtained by the second 
method is not sufficient. Ad- 
ditional reinforcement then 
can be supplied by short bars, 
placed at the support at the 
bottom of the beam. The 
length of the short bars will 








<--- Column 






Direction of Direction of ‘be governed by the bending 
Compression Compression ae 
in Pe ban moment, the minimum length 
Span being such that the projections 
Fic. 97.—Compression Stresses in Bar at beyond the column faces on 
Support. (See p. 284.) each side are equal to the 


length of imbedment required 
by bond for the expected compression stresses. The ratio of com- 
pression steel, p’, is then larger than the ratio of tension steel, p1. 
Depth of Beam Increased by a Haunch.—Another method of 
providing additional compression area is by increasing the depth 
of the beam at the support by a shallow haunch. Under ordinary 
conditions, this method is less desirable than increasing compression 
area by compression reinforcement. While material is saved, the 
excess cost of forms is apt to overbalance this saving. In many 
cases also, the haunch is objectionable from the architectural stand- 
point. A haunch is only useful, therefore, where it is desirable to 
avoid excessive compression reinforcement, or where more lateral 
stiffness is required for the structure. The required depth of haunch 
may be obtained by trial. A depth is assumed and then, with the 
aid of the diagrams on pp. 292 to 297, the stresses are computed 
and a new depth tried out if necessary. The depth is satisfactory 
if the compression stresses do not exceed the specified working 
stresses. 
The depth of haunch may also be found from the rectangular 


PROCESS OF DESIGNING CONTINUOUS BEAM 285 


beam formula (Formula (1), p. 204). In this case, the effect of 
compression reinforeement may be neglected, because for deep 
haunches the bars are placed some distance from the extreme fiber. 
In any case, the error is on the safe side. 

Under ordinary conditions, the computation need be made only 
for the maximum bending moment. The point at which the slope 
should be ended can be readily found from the following formula: 


For a uniformly loaded beam, let 
M, = negative bending moment next to the support; 
M, = moment of resistance of the inverted T-beam without the 
haunch, governed by the concrete; 
a = length of haunch measured from face of column; 
1 = clear span of beam. 


Then 


I 





— M, l 
ie r 11 
2 ea (approximately). Evie tee Reaetey or) 


This formula errs slightly on the safe side. 


PROCESS OF DESIGNING CONTINUOUS BEAM 


_ After the floor is laid out and the spacing of beams fixed, the slab 
is designed first. 

Next the dead and live load per lineal foot of beam is computed 
by multiplying the sum of dead and live load by the spacing of 
beams and adding the estimated dead load of the beam. 

Knowing the condition of restraint at the ends, the shears and 
bending moments of the various spans are computed. | 

The depth of the beam is then decided upon. In a continuous 
beam of equal spans, it is desirable to use the same depth for all 
spans although the bending moments in the end spans are larger 
than in the center spans. 


This formula is based upon the fact that the point of zero moment is at 
approximately 4 of the span. The variation in the moment between the support 
and the } point is assumed to be a straight line. (This assumption is correct 
enough for the purpose.) Hence, the difference between the bending moment 
and the moment of resistance is in approximately the same ratio to the bending 
moment as is the ratio of the distance from the point where the haunch is needed 
to the point of zero bending moment. When the point of zero moment is not 
approximately at 1 span, the fraction may be altered accordingly. 


286 REINFORCED CONCRETE DESIGN 


When the beam is rectangular, the problem is simple. The depth 
is determined for the largest bending moment by Formula (1), p. 204, 
and for the largest shear by Formula (86), p. 247. 

For a T-beam, the problem is solved as aN for T-beam 
design on p. 217. 

The selection of the proper depth for a continuous beam is some- 
what more complicated than for a single beam because the depth 
must be satisfactory for all spans. 

Experience teaches that in most continuous beams, the depth is 
governed by the compression stresses at the support. The proper 
solution then would be to select the ratio of compression to tension 


J 
steel, a and then find the required depth for the largest negative 
1 


bending moment as described on p. 237. It is desirable, if possible, 
to arrange the reinforcement as described in “ First Method” on 


, 
p. 283, in which case the ratio i = Qa: 
1 


If, in a series of beams, a few beams are. subjected to a dispro- 
portionately large bending moment at the support and if, in spite of 
this, it is desirable to retain the same depth for all beams, it will be 
found economical to use a large amount of compression reinforce- 
ment for these few beams. Considered by itself, such a design would 
not be desirable or economical. Taking the series of beams as a 
whole, however, the design will be more economical than if a deeper 
beam were used throughout. 

Having tentatively decided upon the depth of beam and width 
of stem, diagonal tension stresses are computed to see that they do 
not exceed the allowable working stresses. 

After the depth of beam is decided upon, the areas of reinforce- 
ment required in the different spans of the beam for negative and 
positive bending moment are computed, as discussed below. For 
main reinforcement, such bars are selected as will satisfy the positive 
bending moment in each span. Before these bars are finally accepted 
bond stresses should be investigated. 

Bending of bars is then decided upon, and reinforcement for 
negative bending moment investigated. 

After the points of bending of bars is decided upon, the 
required web reinforcement must be designed for beam in which 
the shearing unit stresses exceed those permitted for plain 
concrete. 


PROCESS OF DESIGNING CONTINUOUS BEAM 287 


Tension Steel in Center of Span.—The size and number of bars 
necessary to make up the required cross-sectional area are selected 
for each span independently. If possible, bars of one diameter, or of 
diameters that do not differ by more than 4 in. should be used. It is 
desirable to use not less than two bars in small beams and not less 
than four bars in larger beams. If a larger number is required, bars 
of the largest diameter permissible by bond should be selected, as 
this reduces the number of bars and therefore the cost of handling. 
The number of bars should be chosen, also, to permit the required 
proportion of bars to be bent. up. When bending of one-half of the 
bars is contemplated, an even number of bars should be used. 

Whenever possible, the bars should be placed in one layer. Often 
it is cheaper to widen a beam sufficiently to have all bars in one 
layer than to introduce two layers of bars and thereby increase the 
the depth of beam. With two layers of bars, a strip of concrete, the 
width of the beam, and equal in depth to the diameter of bar plus 3 in., 
is wasted. This may be more than is required for widening the beam. 
In addition, with wider beam the extra concrete is used to resist 
shear, thus reducing the number of stirrups. 

Bending of Bars is Now Decided upon.—The bars should be 
bent with due regard to the bending moments. ‘The points of bend- 
ing should be far enough from the center of the span, where the 
bending moment has decreased sufficiently to permit the reduction 
of reinforcement. The location of points of bending is not the same 
for uniformly distributed loads as for concentrated loads. 

Diagrams, pp. 292 to 297, give the location of points of bending. 
Before using the diagrams, read the notes carefully. | 

In small beams, bars should be bent at one point only at each end. 
Many constructors favor bending the bars at one point only for 
beams of all sizes, because it reduces the number of types of bar on 
the job and thereby facilitates construction. The authors disagree 
with this practice and recommend bending bars in larger beams at 
several points. The disadvantage in the field is more than balanced 
by better distribution of reinforcement in the beam and by the 
increased effectiveness as diagonal tension reinforcement. If a large 
number of closely spaced bars are bent up at one point, they form ina 
beam an inclined plane which impedes the flow of concrete. This 
tends to form a plane of weakness, especially when the bars are not 
kept exactly in position. Proper distribution of the points of bend- 
ing prevent such conditions. 


288 REINFORCED CONCRETE DESIGN 


When bending bars, there is a natural tendency to bend them 
symmetrically in the cross section. While symmetrical arrange- 
ment in a section is preferable, it is not absolutely necessary. Bend- 
ing bars in two points with unsymmetrical arrangement in cross sec- 
tion is usually preferable to bending all bars at one point. 

Tension Steel at the Support.—After the number and the size 
of the bars required in the center of the span are decided upon, the 
requirement at the supports is studied. - 

If Maximum Positive and Negative Bending Moments are Equal.— 
If the negative bending moment at the support and the maximum 
positive bending moment in the center of span are equal, and the 
beam consists of a number of equal spans and is of uniform depth, 
the arrangement of steel is simple. The positive reinforcement is 
designed so that it can be divided into two equal parts. Then one- 
half of the bars are laid straight and the other half bent up and 
carried across the support into the adjoining span to resist one-half 
of the tension there. An equal amount of steel is brought across 
from the adjoining span into the span under consideration. The 
sum of the effective bars at the support is then equal to the sum 
of the bars in the center. The requirement as to the amount of 
steel is satisfied. The only problem is to decide upon the points of 
bending of bars. 

The condition is somewhat different at the first interior support. 
Assuming again that maximum positive and negative bending 
moments in a span are equal, the bending moment at the first interior 
support is equal to the positive bending moment in the center of 
the end span, but is 20 per cent larger than the bending moment in 
the center of the interior span. If the depths are equal, the required 
area of steel is proportional to the bending moments. When one- 
half of the steel in the end span and one-half of the steel in the interior 
span are bent up and carried across the first interior support, the 
available area of steel for the negative bending moment will be about 
10 per cent less than required. ‘The difference must be supplied by 
extra bars of proper length placed in the top of the beam at the 
support. 

At the wall support, where negative bending moment equal to 

2 
a is required, those bottom bars which are bent up, and which are 
half the total number of bottom bars, will not give sufficient reinforce- 
ment for the negative bending moment; neither will the reinforce- 


PROCESS OF DESIGNING CONTINUOUS BEAM 289 


ment extend far enough from the column to take care of the tensile 
stresses at all points. It is necessary to add there, in the top of the 
beam, short bars provided with a hook at the support and extending 
to a point distant 0.27 from the edge of the support. 
If Maximum Negative Bending Moment is Larger than Maximum 
Positive Moment.—For continuous beams running into columns, 
wl? 


where the positive bending moment is 6 while the negative bending 


pol , 
moment is 75 to supply the required amount of tension reinforce- 


ment at the support, two-thirds of the bottom steel must be bent up 
and carried across the support. In other respects, the condition is 
the same as in the previous case. 

It may be found that with such an arrangement the straight bars 
are not sufficient to take care of the compression stresses. In such 
ease, it is a good practice to bend up a smaller proportion of bars. 
and supply the deficiency in tensile reinforcement at the support by 
extra short bars. 

If no Bars are Bent Up.—Sometimes, it is not desirable to bend 
_ up any reinforcement, but to use only straight bars for positive and 
negative bending moment. This method is used in short, deep beams 
where bent-up bars would not reach the top at the proper place to 
be effective as negative bending moment reinforcement. The 
positive bending moment reinforcement is then made up of straight 
bars, some of which extend the full length of the beam while the rest 
extend on both ends beyond the points of inflection. It is not per- 
missible to stop any bars short of the point of inflection, unless the 
bars are provided with a hook of proper dimension to develop the 
existing stress in bars at the point where the bar was stopped short. 
Otherwise there would be no means of inducing any stresses into the 
short bar without slipping of reinforcement. 

The top reinforcement at the support consists of short bars of 
proper area. These should extend on both sides of an intermediate 
support beyond the points of inflection. At the wall column, one 
end should be hooked at the wall and the other should extend beyond 
the point of inflection. 

A design in which the spans are not equal must be studied with 
more care. If the difference in span is large, the bending moments 
must be determined as explained in Volume III. In this case, it 
will not be possible to make up the area of tensile reinforcement at 


290 REINFORCED CONCRETE DESIGN 


the support by the bent-up bars, and additional top bars will be 
required. 

Top Bars to Hold Stirrups.—To keep stirrups in position, two 
straight bars of small diameter are placed at the top, as shown in 
Fig. 81, p. 249. The stirrups are hooked on these bars and wired 
to them. With such an arrangement there is some assurance that 
the stirrups will not become disarranged in process of construction. 


POINTS AT WHICH HORIZONTAL REINFORCEMENT SHOULD BE BENT 


In a beam, the bending moments decrease from a maximum to 
zero at the points of inflection in continuous beams (and at points of 
support in simply supported beams); therefore, the required amount 
of reinforcement at intermediate points, for a beam with constant 
cross section, decreases proportionally. Consequently, it is per- 
missible to bend up a part of the positive moment reinforcement 
at proper points and utilize it either as diagonal tension reinforce- 
ment or as negative bending moment reinforcement. 

Ag the reduction of the required amount of steel depends upon the 
reduction of bending moments, the bending moment curves should 
be used as a guide in preparing bending sketches for the bars. To 
facilitate the work, bending moment diagrams for uniformly dis- 
tributed loading are given on pp. 292 to 297 for different conditions 
of restraint at the support. Under each moment diagram, corre- 
sponding bending details are given. It should be noted that in 
the bending moment diagram the maximum bending moment is 
assumed as a unit and the intermediate bending moments are ex- 
pressed as a fraction of the maximum moment. Where the positive 
and negative bending moments are equal, the assumed units for 
both eases are equal. Where the maximum positive bending moment 
is different from the maximum negative bending moment, the mag- 
nitudes of the units are also different. 

To facilitate the use of the diagram, horizontal lines divide the 
maximum bending moment into ten parts. The span is also divided. 
In this way, the bending moment at any point may be found from 
the diagram in terms of the maximum moment. 

In all diagrams for continuous beams, the distance between 
the supports was assumed as the span, in accordance with the rec- 
ommendation given on p. 277. The diagrams may be used for other 
assumptions as to the length of the span. However, if for any 


POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 291 


reason it is required to assume the point of support to be within the 
column, the distance of points of bending of negative reinforcement 
should be based on the net span and measured from the edge of sup- 
port (and not from the theoretical point of support). 

Explanation of Max. and Min. Dimensions in Bending Diagrams. 
—In the bending diagrams for each bend, two hmiting dimensions 
are given, one at the bottom and the other at the top. ‘This does not 
mean that each bent bar will be bent at both these points. The two 
points are limiting points. 

The bottom dimensions are given as, say, Max. 0.251. This means 
that the distance of the bent of the bar from the edge of the support 
must not exceed 0.251. As far as positive bending moment is con- 
cerned, however, the bent may be moved nearer towards the support 
and the distance of the bent from the support may be smaller than 
the allowable maximum value. 

The top dimensions are given as, say, Min. 0.15/. This means 
that the bent must not be placed nearer the support than 0.151. 
As far as negative bending moment is concerned the position of the 
top bent may be moved away from the column. 

Any bend between these points is satisfactory. In selecting 
the actual points of bending, the diagonal tension should also be con- 
sidered. In fixing the points of bending, it is always advisable to 
take into account the possibility of small inaccuracies in bending of 
bars and provide for it by placing the points of bending a few inches 
from the theoretical point. Thus, the point at the bottom should 
be a few inches nearer the support, and the point at the top a few 
inches farther from the support, than the theoretical values. 

In practice, the point of bending at the bottom will be assumed, 
and the bar bent at an angie of 35 to 45°. The point of bending at 
the top is obtained automatically. If the distance from the support 
to this top point is equal to or larger than the minimum given in the 
bending diagrams, the design is satisfactory. If the point at the 
top comes too near the support, a steeper angle should be used. 
Although the angles recommended above are desirable, as then the 
bar is most effective as web reinforcement, angles of any steepness 
may be used if necessary. 

The use of the diagrams is shown in Example 1, on page 578. 


292 REINFORCED CONCRETE DESIGN 











(4M, 7M, : 
jie Ot 
_ 0.3 M,.. 1 
= 0.2 M; a|S 
Ae ae 
Se 01M & 
eee 6.20 








Bending. Moment Curve 
ns OLS 12 4 SS 


—_—--- ~~ 


A,=Total Area 
Pos. Reinf. 


omar. 0.221 --------- 





Bending Details 
Fic. 98.—Beam Freely Supported at Both Ends. (See p. 290.) 


Note: For formula of bending moments see p. 278. 
For explanation of terms Max. and Min. used in bending diagrams see p. 291. 
The limiting dimensions for bending do not always coincide with selected points for 


bending of bars. 


POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 293 



































posal bone de 
EG PIS OE) EME us 
mer 







0.151412; Bending Moment Curve 
seencb nce Weg. Reinf=Ag at Right ee. -28 










A,= Total Area Pos. Reinf. 
Trax pom eS dary 
Bars Bent at One Point 
~Brick 
Neg. Reinf=A, at Right 0.21412 














1 
‘ SA 0.09A,. 






A,=Total Area geseteeteen sy 
ha Pos. Reinf. e-Mae 028 
~Max. 0,131 1 


ee er eee ee ere ee ree 







Bars Bent at Two Points 


- ‘Brick 


Bending Details 
Fic. 99.—End Span of Continuous Beam, Freely Supported at End. (See p. 290.) 


Note: For formula of bending moments see p. 279. 
For explanation of terms Max. and Min. used in bending diagrams see p. 291. 
The limiting dimensions for bending do not always coincide with selected points for 
bending of bars. 


294 REINFORCED CONCRETE DESIGN 


































pan 
2 











“a 


Miki 








Bending Moment Curve 









Min. 0.101------ <0. St Neg. Reinf.=A, each End |<0.21+12- 
Min. 0.051 : 


At As | 4A,-~- 















Bars Bent at 


Two Points -- Max. 0.3l-- A xTotal Area of Pos. Reinf. Nee Point 


RN ae Re See a 


a yor 
Bars Bent at Ee 0.21 


Bars Bent at Three Points 


Bending Details 


Fic. 100.—Interior Span of Beam Continuous for More than Two Spans. 
(See p. 290.) 


Note: For formula of bending moments see p. 279. 
For explanation of terms Max. and Min. used in bending diagrams see p. 291. 
The limiting dimensions for bending do not always coincide with selected points for 
bending of bars. 


POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 295 




















on VL eee 0.551 : 
Do Site 
= AS ie. | SP Re ae 
ee COE ee I 
1 | OUT BS ie i ae ee ee 
Pun SE 2a 
“seo Ah sa 
SIS pea se Tah Sled st 
J Se Be Bs 
UR LI ee Be : 
S Nisioer 1 1) | 
Se Oma ae 
ude! Sib. ea eters 
es a ES | 
oN ae ck) 
maAN Pia 
= 
3 
, S 


E - m2 = ----- - --------- -----] ---- -------- 


Bending Moment Curve 
Neg. Reinf. =A, at Right 
0.7A, at Left. 





_ars Bent at One Point 


ke.-9 Q]->4 Neg. Reinf.=A , at Right yr 0.212: 
m OD Alect letter ear 









Max ae A, = Total Area 
Pos. Reinf. 






0.25; 
<- Mas. 0,321 


Bars Bent at Two Points 


Bending Detatls : 
Fic. 101.—End Span of Building Frame End Partly Restrained. (See p. 290.) 


Note: For formula of bending moments see p. 280. 
For explanation of terms Max. and Min. used in bending diagrams see p. 291 
The limiting dimensions for bending do not always coincide with selected points for 
bending of bars. 


296 REINFORCED CONCRETE DESIGN 




























NEGIMA EST SRI a0, 
CHE eae 
REBUMIADE GSES 
Ne Ra ged So eee 
Srey MANES Be a eel | ee 
oy SEA a Sols be a mmbety 
1 ENE 
ss DNS) eee a 
Bl ee SAAS 
PEN afer) an Se 
Pe ep 
Lo Vee a 
Ae Sees 
Ni ee 
Pe I AW a Gea Py 
SG GNM mile 
aay SS Rey 
me BLE ee 
GEN Es har FN waele 5, 
Ses muy 





~en maw eer eee am — eens == 


Bending Moment Curve 





0.21412 


As=Total Area Pos. Reinf. \_Ma : 


witene-- +--+ --------------] ecw ewan cana ness s- ses ee 


Bars Bent at One Point 


0.21412" 






0.21412" 




















~+--6 


“ Neg. Reinf=A, Each End 
“>f~--~4 Min.0.101 Min.0.101-7 
: Min. 0051-4" 


---> 
-7 = 


a 
-- 5 








ween Tage NOX 0. ‘As 
A =Total Area Le-Max. 0.31---- 
_ Pos. Reinf.___._. 


ee ee D 


Bending Details 
Fic. 102.—End Span of Building Frame. End Fully Restrained. (See p. 290.) 


Note: For formula of bending moments see p. 280. 
For explanation of terms Max. and Min. used in bending diagrams see Pp. 291. 
The hmiting dimensions for bending do not always coinciae with selected points for 


bending of bars. 


POINTS AT WHICH REINFORCEMENT SHOULD BE BENT 297 


. 























== 
ane 
: 



































Bending Moment Curve 
0.21 + 12" Neg. Reinf.=1 3A, each End 
Str. Bar <~-Min. 0.071 ; 
gd Min. 0.081 








t int 
Pare Pea Ot Tie foints Bars Bent at One Point 


One - Half Positive Reinforcement Bent 


» Neg. Reinf. =1 +A, each End 
0.21 + 12> PIR cha 












A, = Total Area Pos. Reinf, | Max. 0.21 


----2----->| 


> 






Bars Bent at Two Points Bars Bent at One Point 


Two-Thirds Positive Reinf. Bent 
Bending Details 
Fic. 103.—Interior Span of Building Frame. (See p. 290.) 


Note: For formula of bending moments sce p. 279. 
For explanation of terms Max. and Min. used in bending diagrams see p. 291 
The lmiting dimensions for bending do not always coincide with selected points for 
bending of bars. | 


298 REINFORCED CONCRETE DESIGN 


REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE STRESSES 


All masonry is subject to temperature cracks, but when they are 
distributed in the many joints between bricks or stones they do not 
show so plainly as on the smooth surface of concrete. 

Expansion due to a rise in temperature rarely causes trouble 
except at angles where the lengthening of the surface may produce 
buckling or sliding of one portion of the wall past the end of the 
other. Ina building, the walls and floors are generally so well bonded 
together and free to move as a unit, that no provision need be made 
for expansion. In a structure like a square reservoir, the effect of 
expansion must be taken into account in the design, to prevent failure 
at the corners. 

Contraction is often more serious, although cracks are by no 
means necessarily dangerous. ‘Temperature reinforcement is ordi- 
narily used to prevent excessive cracking due to the shrinkage of the 
concrete in hardening or to the lowering of the temperature (see 
Volume II. Also, expansion joints are used in long structures. 

Temperature reinforcement, placed in the direction of the expected 
contraction, does not prevent cracking entirely; but it distributes 
the cracks in such a way that numerous very small cracks, which are 
often invisible to the eye, are formed instead of a few large ones. 
To accomplish this purpose, the temperature reinforcement must be 
sufficient in quantity, and it should consist of bars of small diameter 
placed as close as practicable to the surfaces. Deformed bars, that 
is, bars with irregular surfaces which provide a mechanical bond with 
the concrete, are more effective in distributing cracks than smooth 
bars. Steel of high elastic limit also is advantageous. 

The area of temperature or shrinkage reinforcement used in 
practice varies from 0.2 to 0.4 of 1 per cent of the cross section of the 
concrete (p = 0.002 to 0.004). ; 

The amount of temperature steel to be used depends upon con- 
ditions and the degree to which the cracks should be reduced. Ifa 
water-tight job is required, the higher value should be used. 

The tensile strength of concrete is so low that a small change in 
temperature will crack it. For example, the coefficient of expansion 
of concrete is 0.0000055 (see Volume IT) and the modulus of elastic- 
ity is generally assumed as 2,000,000; therefore, the stress per 
degree Fahrenheit is 0.0000055 X 2,000,000 = 11. Ib. peresanaie, 


REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE 299 


and a fall in temperature of 32° = 27° is sufficient to crack a con- 
crete the tensile strength of which is 300 lb. per sq. in. 

It is evident, and it has been proved by experience, that there is 
less cracking in concrete laid in cold than in warm weather. 

Longitudinal reinforcement is especially necessary in conduits, 
which must be water-tight. 

Cracks due to shrinkage in hardening of the concrete may be pre- 
vented by keeping the concrete wet. (See Volume II.) 

It has been suggested by Mr. Charles M. Mills that the relation 
between the tensile strength of the concrete and the bond with the 
bars is an important factor in governing the size of the cracks; and 
the following analysis, based on his suggestions, gives a means of esti- 
mating the size and distance apart of the cracks, thus forming a basis 
for judgment as to the sizes and percentage of steel to use. 

The tensile stress in the steel at a crack tends to pull out the bars 
from the concrete, and, referring to Fig. 104, the bond stress of the 
bar in the length ab must equal the tensile stress in the whole cross 
section of the concrete at b caused by the contraction of the concrete. 


Let x = distance apart of cracks inches; 
D = diameter of round bar or side of square bar; 
p = ratio of cross section of steel to cross section of concrete. 


Then,!2 if, as is sufficiently accurate for practical purposes, the 
strength of concrete in tension is assumed to be equal to the bond 
between plain steel bars and concrete, the distance apart of cracks is 


CL = ; “ for square orround bars. . . . (76) 


The distance apart is inversely proportional to the unit bond,” 
so that a deformed bar having twice the bond strength would space 


12 In addition to above notation, let 


A = area of section of concrete; u = unit bond between plain steel and 
A; = area of section of steel; concrete; 
o = perimeter of steel bar; f,; = unit tensile stress in steel; 
fi = tensile stress in concrete; D = diameter of bar. 
2Af 2A 
Then Af; = juor, or x = =! If fp =u, 2 = —, and since p = —, 
uo 0 a. 

2A A D 1D 
~=-——. Also, — = — for both round and square bars, hence x = 5 —. 

op O 2p | 


300 REINFORCED CONCRETE DESIGN 


the cracks one-half as far apart and allow them to be only one-half as 
wide. 

_ It is evident that the distance apart of the cracks is proportional 
to the diameter of the reinforcing bars, and inversely proportional 
to the percentage of steel. 


a 6b 









Fig. 104.—Reinforcement for Temperature Stresses. (See p. 299.) 


From this formula is tabulated the estimated percentage of rein- 
forcement for different spacing of cracks and different sizes of bars, 
assuming the bonding strength of the steel to the concrete to equal 
the tensile strength of the concrete. 


Estimated Percentage of Reinforcement for Different Spacing of Cracks 





Distance Apart of Cracks 




















Pisin Barsey so. oso Ve 18” 24% 36” 48". 60”’ 
Deformed Bars *..... 8” jee 16” 24” oo 40” 
% % % % % % 

1) 1.04 0.70 0.22 0.35 0.26 0.21 

Diameter of round | 2”) 1.56 1.04 0.78 0.52 0.39 0.31 
or side of square | 3”) 2.08 1.39 1.04 0.69 0.52 0.41 
bar: Ae Bie 5" 2.60 1.74 1.30 0.87 0.65 0.52 
Boeke 2.08 1.56 1.04 | 0.78 0.62 

i”) 3.65 2.44 1.82 122 0.91 0.73 

Pe er. 1.04 0.83 








2.78 2.08 Ber: 








Norte: To express the steel as the ratio of area of cross section of steel to cross section of 
concrete, divide the percentages by 100; thus 1.04 becomes p = 0.0104. 


* Assuming the bond of deformed bars to be 50 per cent greater than plain. 


The size of the crack is governed by the amount of shrinkage and 
for cracks due to temperature changes may be estimated as the 
product of the coefficient of contraction (0.0000055) by the number 
of degrees fall in temperature by the distance between cracks, 


REINFORCEMENT FOR TEMPERATURE AND SHRINKAGE 301 


Estimated Width of Cracks for Different Distances Apart 


Distance Apart 





Change of 
Temperature 
ja 18” 24’ 36” 48’ 60’ 
CT OSL SE ited ir 0.0020 | 0.0080 | 0.0040 | 0.0059 | 0.0079 | 0.0099 
ae ee 0.0033 | 0.0050 | 0.0066 | 0.0099 | 0.01382 | 0.0165 
CO SD i ora 0.0046 | 0.0069 | 0.0092 | 0.0139 | 0.0185 | 0.0232 


* 30° corresponds to a shrinkage of 0.017 per cent; 50° to 0.028 per cent; 70° to 0.038 per 
cent. 


From this, if it can be determined how large a crack will be allow- 
able, the corresponding spacing can be obtained. 

To avoid large cracks it may be necessary to use enough steel to 
prevent its passing its elastic limit. If the bars are continuous for 
such a length that the ends are practically immovable, as in a long 
retaining wall, a drop in temperature, tending to shorten them, pro- 
duces a tensile stress which is independent of the distance between 
the restrained ends. Assuming the coefficient of expansion of steel 
to be the same as that of concrete and the modulus of elasticity of 
steel as 30 000 000, this stress is 30 000 000 X 0.0000055 = 165 Ib. 
per sq. in. per degree of temperature, or for 50° F. is 8 250 lb. per 
sq. in. This is well within the elastic limit of the steel and would 
not, of itself, cause the steel to take a permanent set. However, 
since the concrete surrounding the steel will be continuous except at 
certain cracks, the stretch in the steel may be unevenly distributed 
and largely confined to the immediate vicinity of the cracks. If 
cracks occur while steel is unstressed, through the concrete shrinking, 
the steel tends to resist the shrinkage by tension at the crack and 
compression at the center of the block of concrete, and the tensile 
stress will be equal to the compressive and each equal to one-half 
the tensile strength of the concrete. This may be expressed by the 
following formula, in which the foregoing notation is used.!3 


1 
fe = gfe 
13 Af, : fad, A ; toy 15 
iil on NR Pore Maaco 


302 REINFORCED CONCRETE DESIGN 


See the tensile stress in the concrete is liable to be low at the 
time shrinkage cracks are formed, it may be assumed, for illustration, 
as 200 Ib. per sq. in., making 


LO) 
I's a 

This represents the stress due to local cracks, which is additional 
to the temperature stresses above described. The total stress is, 
therefore, for 50° change of temperature 8 250 +- f’, or 8 250 + 
If the clastic limit of the steel is 40 000 Ib. per sq. in., and we must 
_ keep below this, 


40 000 = 8 250 + - and p = 0.0031. 
For steel, the elastic limit of which is 50 000 Ib. per sq. in., 


50 000 = 8 250 + ~" and p = 0.0024. 


These values of p represent the lowest theoretical ratio of area of 
cross section of steel to area of cross section of concrete which can be 
used without the steel passing its elastic limit at certain of the cracks 
when the ends are restrained or the length is so great that inter- 
mediate parts are practically restrained. 

In view of the very slight stretch required to relieve the stress in 
the bars when the elastic limit is exceeded, and the probability of its 
distribution by the restraint to movement by the mass, it is not 
always essential to consider the elastic limit. 


CHAPTER VI 
DESIGN OF FLAT SLAB STRUCTURES 


The aim in this chapter is to present a complete and logical treat- 
ment of the design of the flat slab. Precedent and common practice 
are followed when they agree with sound engineering. A number of 
rules of thumb, however, which have accumulated during the early 
stages of flat slab development have been eliminated, and rules 
based on logic and on accepted principles governing other types of 
concrete design have been substituted. 

In this chapter, besides covering the general principles of flat 
slab design, the authors aim to treat the details in such a way as to 
give the engineer, and also the student, the basic information and 
instruction required to design a safe and economical slab adapted to 
any required conditions. Design of interior columns is treated on 
p. 305; that of exterior columns on p. 312; column heads and drop 
panels, p. 319; rules for thickness of slab, pp. 325 and 337 ; bending 
moments to be used in design, p. 328; formulas for compression 
stresses in slab, p. 341; use of compression reinforcement in flat 
slab design and formulas, p. 345; shearing stresses in flat slab, p. 
347; method of computing tensile reinforcement and arrangement 
of bars in flat slab, p. 351. 

For computing bending moments in flat slabs, the authors have 
adopted the widely accepted method based on the division of the 
slab into parallel strips. This permits a fairly well balanced design 
which, with proper tables, can be worked out without excessive labor. 
For a method more logical in theory and still simpler in practice, the 
reader who desires to investigate theory is referred to the treatment 
in the Second Edition of this book, which was worked out by the 
authors. The strip method, however, has become established by 
precedent and gives workable results; for the present, therefore, it is 
accepted by the authors in preference to introducing a different type 
of formulas. Further experiment and analysis will probably throw 
new light on the subject. 

303 


304 DESIGN OF FLAT SLAB STRUCTURES 


The formulas for bending moments, recommended for use, agree 
in general with those given by the 1924 Joint Committee, which, in 
turn, correspond with those now in common use. Modifications 
have been made by the authors, however, where necessary for uni- 
formity of design. The formulas may be safely used for rows of 
equal or substantially equal panels. Unusual designs should be 
analyzed according to formulas given in Vol. III. 

In all respects, the material in the chapter has been based upon 
the considerable experience of the authors in design of flat slab con- 
struction, and upon various tests, many of which were made by or 
for the authors. 

Advantages of Flat Slab Construction.—In flat slab construction, 
concrete is used more logically than in beam and girder construction. 
Whereas, in beam and girder construction, the load is transmitted 
from the slab to the beam, then from the beam to the girder, and 
finally from the girder to the column, in flat slab construction the 
load is transmitted directly by the slab to the column. The load, 
irrespective of its position on the slab, is carried by the whole slab; 
therefore, overstressing, due to concentrations of load, is impossible. 

The under side of a flat slab is practically flat. Because of the 
absence of exposed corners, flat slab construction is less vulnerable 
in case of fire than beam and girder construction. 

The formwork for flat slab 1s simple. Forms for column heads 
and drop panels, usually made of metal, are adjustable and can be 
rented at a small cost. - Ordinarily, scarcely any adjustment of forms 
is necessary for different floors designed for different loadings, since 
the change in thickness of slab does not require change in formwork. 
Change in thickness of drop can be readily accomplished. 

Universal Adoption of Flat Slab Construction Owing to its 
many advantages, flat slab construction has practically superseded 
beam and girder construction for spans Up to 30 ft. and live loads 
over 100 lb. per sq. ft. For smaller live loads, light-weight floors are’ 
often more economical. For more detailed discussion, see Chapter 
and also discussion of various types of buildings. 

Description of Flat Slab Construction.— Reinforced concrete flat 
slab construction is distinguished from all other types of slabs by 
the absence of beams and girders. The slab carries the load and is 
supported directly by columns, which are provided ordinarily with 
enlarged heads, called column heads or capitals. For heavy loads, 
the slab is thickened at the column by drop panels or plinths. The 


COLUMNS IN FLAT SLAB CONSTRUCTION 305 


_ elements of flat slab construction are, therefore: columns, column 
heads, slabs, and drop panels. For light loads, the drop panels, and 
sometimes even the column heads, are omitted. The functions of the 
elements are discussed under separate headings. 

Flat slab construction is illustrated by photographs in Figs. 105 
and 106. In Fig. 105 is shown light flat slab construction without 
drop panels. The pleasing appearance of the room, which is used 
as an office space, is noticeable. Attention is called also to the dis- 
tribution of the light. Fig. 106 shows a heavy flat slab construction 
with drop panels, as used in a warehouse. 

Reference is also made to the photograph of flat slab building, 
Fig. 185, p, 572, and also to the perspective view, Fig. 187, p. 574. 


COLUMNS IN FLAT SLAB CONSTRUCTION 


In flat slab construction, distinction is made between interior 
and exterior columns, not only on account of their shape but also 
because of the difference in function. 


INTERIOR COLUMNS. 


Interior columns may be round, square, octagonal, rectangular, 
or oblong in cross section, as required. In most structures, the 
only consideration affecting the shape of the column is economy 
of the formwork for the column and the column head. Where 
metal forms are easily obtainable, round columns and round column 
capitals are most economical, especially with spirally reinforced 
columns. Octagonal or square columns are sometimes specified. 
Octagonal columns are considered by some architects to be more 
pleasing in appearance than round columns. Square and _ octag- 
onal columns are also preferred, in some cases, because it is easier 
to fit partitions to the flat surfaces of the columns. Square and 
octagonal columns are also used where metal forms are not readily 
obtainable, since wood formwork for either square or octagonal 
columns and column heads is cheaper than for round columns. 

Rectangular or oblong interior columns are used where it is desir- 
able to keep the clear span in one direction as large as possible. Such 
may be the case in manufacturing establishments, such as shoe 
factories and textile mills, in which the manufacturing processes or 
machines run in one direction, and a strip of the building of the width 


306 








DESIGN OF FLAT SLAB STRUCTURES 


Fic. 105.—Interior of Office Building. (See p. 305.) 


Densmore LeClear and Robbins, Architects. 8. M. I. Engineering Co., 
Consulting Engineers. 


Fic. 106.—Interior of Warehouse. (See p. 305.) 
Chas. H. Way, Architect. §S. M. I. Engineering Co., Engineers. 


COLUMNS IN FLAT SLAB CONSTRUCTION 307 


of the column is unoccupied space. Rectangular columns are some- 
times used in garages and in printing establishments, to make more 
room for the machines between the columns. 

Design Formulas.—lormulas for design are given in the chapter 
on Column Design. The economy of design of the columns is also 
discussed there. 


INTERIOR COLUMNS FOR SYMMETRICAL ARRANGEMENTS OF PANELS 


When the arrangement of panels is symmetrical about the center 
line of the column and the load is uniformly distributed over the 
whole area, the interior column is subjected to vertical load only, 
and no bending moment acts in it. However, in the actual use of 
the building, occasional unbalanced loading is unavoidable. This is 
caused by live load placed in panels on one side of the column line, 
while the opposite side is not loaded. Such loading produces bending 
moments in the interior columns. Ordinarily, the possibility of 
such bending moment is taken care of by limiting the minimum size 
of the interior column to a definite ratio of the span. The columns 
are then designed for the vertical load only. 

Limitation of Minimum Size of Interior Column.—Most specifi- 
cations and building codes require that the diameter of the interior 
round column should be not less than a certain fraction of the span 
measured center to center of columns. The requirements of various 
building codes are given in the table on p. 308. 

Authors’ Recommendations.—The authors recommend the fol- 
lowing limitations of interior columns: 


‘ Limitation of Outside Diameter 


Design Live Loads, Ib. per sq. ft. of Interior Round Column 


60 Ib. and under No limitation 

70 to 100 Ib. zis of span 
110 to 200 Ib. z's of span 
210 to 300 lb. 5 of span 


For higher loads, computations of bending moments should be 
made. 

The above limitations apply when the bending moment due to 
unbalanced load is resisted by a column above and below the floor. 
The columns should be provided with vertical bars of an area equal 
to at least 1 per cent of the effective area of the column, but not 


308 DESIGN OF FLAT SLAB STRUCTURES 


less than four 2-in. round bars should be used in square columns 
nor less than six -in. round bars evenly spaced around the circum- 
ference in round columns. If no column extends above the floor, 
the minimum diameter of column should be increased by 15 per cent, 


Limitation of Diameter of Interior Columns for Flat Slabs 
With Symmetrical Arrangement of Panels 


(In Effect in 1925) 

















Cit matte Diameter| Other Diameter Provision for 
y in Terms of Span Limitations Unbalanced Loading 
Bostonians. eaters z's span in longer 
direction 
Chicago ais 0" «3 zis average span zis clear height 
(| =}; average. span 
A Ate 
Cine | é for floors Min diameter, 
qs average span 14 in. 
for roof 
New YOrk, 2. twa. z's average span Min., 16 in. for | Moment of zoW il* to 





round, and 14 be resisted by column 

in. for square above and below in 

a proportion to their ‘ 

Philadelphia. 2.2)? sak 6 04 = 29k 4: Bee ee Interior columns should 
be capable of resist- 
ing unbalanced bend- 
ing moment pro- 
duced by panel with 
live load adjacent 
to a panel without 

| live load 
Joint Committee, 
1924) 3 eee No special |requirement given 








* W , = total live load on panel in lb.; 1 = average span center to center of columns in feet. 
Moment is in ft.-lb. : 


or the amount of steel increased beyond one per cent, in accordance 
with the discussion given below. 

The limitation of the column size serves the purpose of supplying 
sufficient rigidity for the column to take care of the bending moment 


COLUMNS IN FLAT SLAB CONSTRUCTION 309 


produced by occasional unbalanced loading. The limitations given 
are based on experience and are satisfactory with average spans and 
average heights of columns. In special cases, the bending moment 
due to unbalanced live load in the interior column should be com- 
puted in the manner given below. This should be done for unusual 
heights of columns, special sections, and especially heavy live loads. 
Stresses due to bending should be computed when the loading in the 
panels is not uniformly distributed, for instance, if one panel carries 
much heavier machinery than the adjoining panels. 

The authors’ recommendations are based on the following rea- 
soning. 

Since the limitation of the diameter of interior column is for the 
purpose of supplying in it sufficient resistance to bending caused by 
unbalanced loading, the required resistance should depend not only 
upon the span but also upon the magnitude of the live load and upon 
_ the possibility of unbalanced loading. If a column with a diameter 
equal to, say, +51, is rigid enough for a warehouse designed for a live 
load of 150 Ib. per sq. ft., where it may have to resist large bending 
moments caused by one-sided loading of panels, a smaller diameter 
may be used with equal satisfaction for an office or a hotel with live 
loads from 40 lb. to 70 lb. per sq. ft. and where heavy unbalanced 
loading is practically impossible. 

The limitation of the diameter of column to a fraction of the 
span, say ;';l, applies directly to round or octagonal columns provided 
with a minimum amount of vertical steel. An increase in the 
amount of steel increases appreciably the magnitude of the moment 
of inertia. Therefore, if it is necessary for any reason to use a column 


having a smaller diameter than the required rigidity may be 


l 
15’ 
obtained by using an increased amount of vertical steel, so that the 
moment of inertia of the column is equal to that of a column with a 


diameter equal to “ but with the minimum amount of steel. 


If square or rectangular columns are used, the limitations in the 
table do not apply directly. In such cases, the column selected 
should have a moment of inertia equal to that of a round column of a 


diameter equal to the required fraction of the span. The moment of 
4 ° . 
inertia of a square with a side, d, is is or 0.082d*, while for a circle 


with a diameter, di, it is 0.049d;4.. For equal moment of inertia, 


310 DESIGN OF FLAT SLAB STRUCTURES 


4 
= = 0.049di4, from which d = 0.87di. This means that, for equal 
moment of inertia, the side of a square may be only 0.87 of the 
diameter of the round column. When the minimum diameter of round 


column is required to be 7s!, the minimum side of a square column 


1 
should be 17.2" 


For a wide rectangular column equivalent to a round column 
with a diameter equal to ;';/, the thickness of the column, for rigidity, 


may be much smaller than jl, as illustrated below. 

Example of Square and Rectangular Columns.—Tlor a specific example, 
assume a 25-ft. square panel and a limiting fraction j'sl.. The minimum allowable 

25 X12 
15 

-the side of the square may be made equal to 20 X 0.87 = 17.4 in. Now assume 
that for a special reason it 1s desirable to reduce the thickness of the column in 
one direction by using a rectangular column 30 in. wide. The moment of inertia 
of arectangular column, where b is width and d thickness, is 7,bd*. For b = 30, 
this is 39d? = 2.5d%. The moment of inertia of a 20-in. round column is 
0.049 X 204. Comparing the two values, 2 5d3 = 0.049 X 20!, from which 
d= 14.6in. Thus, in this case, a 20-in. round column, a 17}-in. square column, 
or a 14.6 by 30-in. rectangular column may be used. 





diameter of a round column is — 20 in. If a square column is used, 


In a rectangular panel with a round or square column, the min- 
imum diameter is based on the average span. If a rectangular or 
oblong column is used, the rigidity in the long direction may be 
made equal to that of a round column of a diameter equal to the 
specific fraction of the long span, and the rigidity in the short direc- 
tion may be made equal to that of a round column with a diameter 
equal to the same fraction of the short span. , 

The rigidity of the column depends not only upon the moment of 
inertia but also upon the height of the column. Rigidity is expressed 
by the ratio 7 The limitation given above applies to columns of 
ordinary height, say up to 14 feet. The rigidity decreases with the 
increase in height of the column. Therefore, for higher columns, the 
diameter of the column should be increased above 4'5/ if it is desired 
to make its ratio of stiffness equal to the ratio of stiffness of a 14-foot 
round column with a diameter ;'5!. 

Bending Moment in Interior Columns.—Some building codes, 
notably those of the cities of Philadelphia and New York, and the 
1924 Joint Committee recommendations require that interior columns 


COLUMNS IN FLAT SLAB CONSTRUCTION oll 


be computed for the bending moment produced by the live load 
placed on one side of the column with adjacent panels without live 
load. This bending moment may be computed by the following 
formulas, which apply when the column extends above and below 
the floor under consideration. 

Let W;= total live load on the loaded panel, Ib. ; 

! = panel length, ft.; 
M, = bending moment produced by unbalanced live load. 
Then 
Bending Moment due to Unbalanced Live Load, 


1 
M, = 40 W, HR By ee ss ‘ 5 é (1) 
PEON Vapi ith lain wh 2069} 


This bending moment should be considered as resisted by the 
columns above and below the slab in proportion to their ratios of 


I, 
stiffness i: If the stiffness of the two columns is equal the moment 


M 
M to be resisted by each column equals sat When the difference 


between the ratios is appreciable, the proportion of the moment 
resisted by each column should be computed. Usually it is near 
enough to assume that the lower column resists 0.61 and the 
upper column 0.4M;. 

The points of application of the bending moment are evident — 
from Fig. 107, p. 311. 

This bending moment is combined with the column load, and 
stresses are computed by formu- pe 
las for direct load and moment Panel». 
(see p. 163). 

The combined compression 
stresses due to the direct load and _ Tension--f 
the bending moment should not 
exceed the stresses recommended 
on p. 463. Of course, the column 
must be strong enough to resist the 
maximum direct stresses without 
exceeding the stresses specified for Fig: 107:-ePendineemromene: aise 
columns. terior Columns. (See p. 311.) 

If the column steel is not suffi- 
cient to resist the tensile stresses, additional bars should be added. 





Loaded Panel 


312 DESIGN OF FLAT SLAB STRUCTURES 


Since, ina square panel, some bending moment may occur at any 
side of the column, the required additional reinforcement must be’ 
so placed as to be effective on all four sides of the column. It is 
best to place such bars in the corners, since then they may be con- 
sidered as effective for bending moments acting on either of the adja- 
cent sides. In rectangular panels, bending moments in both direc- 
tions may have to be computed. 

It is not necessary, in columns, to consider unbalanced live loading 
due to the roof load. The most unfavorable effect of the bending 
moment on the column will be found in the top story column, where 
the bending moment due to floor load is essentially the same as in 
other floors, but the size of the column and the amount of reinforce- 
ment is smaller than in any other story. 

Unsymmetrical Arrangement of Interior Columns.—For unsym- 
‘metrical arrangement of interior columns, the bending moments in 
columns exist even for uniform loading. The required bending 
moments in columns and slabs should be determined as explained in 
Volume III. 


EXTERIOR COLUMNS. 


Exterior columns are usually rectangular in shape. A rectangular - 
column imbedded in the wall projects less into the building than a 
square column carrying the same load. Also, exterior columns 
often serve as pilasters between windows, in which case their width 
depends upon the width of the window sash. For this and other 
reasons, it is advisable to make the width of the column constant 
for the whole height of the building. 

Some specifications (notably the Flat Slab Regulation of the City 
: : : beeeens: aa) OL Oa ete 
quire large mini- 
mum dimensions 






Typical Exterior 





ee Special Exterior Columns for wall columns. 
Fic. 108.—Cross Sections of Exterior Columns. In such cases, rect- 
(See p. 312.) angular columns 

Note: Reinforcement not shown. may not be eco- 


nomical. The sections shown in Fig. 108, p. 312, are often used 
under these conditions. 

Design of Exterior Columns.—Exterior flat slab panels, when 
supported by concrete columns, are designed as partly restrained at 
the ends. To get this favorable condition for the slab, the exterior 


COLUMNS IN FLAT SLAB CONSTRUCTION 313 


columns must have sufficient rigidity to produce in the slab the 
required restraint. Also, they must have sufficient reinforcement to 
resist the bending moment transferred from the slab to the column. 
The fact that exterior columns are subjected to bending and that the 
strength of the slab in the wall panels depends upon the strength 
and rigidity of the exterior columns cannot be overemphasized. 
Outside columns of light or improper design are the most frequent 
cause of weakness in flat slab design. 

Required Rigidity of Exterior Column.—The rigidity of a member 
depends directly upon its moment of inertia, J, and inversely upon 


its height, h, or length, 1. It is represented by the ratio - for columns, 


and by : for beams or slabs. 


Exterior columns should be made rigid enough to develop in the 
slab, at the wall columns, at least eight-tenths of the negative bend- 
ing moment in an interior panel. By the theory of least work, the 
authors find that for this purpose the relation between the ratio of 
rigidity of the column and that of the slab is expressed in the formula 
below. | 

Let J = moment of inertia of slab; 

1 = length of span perpendicular to the wall; 
I, = moment of inertia of column; 
h = height of column. 


Then . 


mes 


Sy reenter aetna ee eentCy 


From this relation, the required dimensions of the column may 
be computed as given below. The moment of inertia of the slab 
should be based on a thickness equal to the thickness of the slab plus 
one-third of depth of drop panel, and a width equal to the spacing of 
the exterior columns. 

The minimum thickness of a rectangular wall column for a given 
width, 6, determined by the required rigidity, may be easily found 
from the following equations: 

Let J = length of span of slab at right angles to the wall, from 

center to center of columns, ft.; 
1, = length of span of slab parallel to the wall, from center 
to center of columns, ft.; 


314 DESIGN OF FLAT SLAB STRUCTURES 


h = height of column from center to center of slab, ft.; 
b = width of column, ft.; 
d = depth of roof slab or top floor slab, in.; 
d, = minimum thickness of column, in.; 
Then 


I 


dy = 1.154) a rs 


For slab with drop panels, the value d in the above formula equals 
the thickness of the slab plus one-third of the depth of the drop 
panel. 

Bending Moment in Slab at Wall Column:—The magnitude of the 
negative bending moment in the slab at the wall column, which is 
transferred to the columns, depends upon the relative rigidity of the 
column and the slab. This bending moment should not be taken as 
less than eight-tenths of the negative bending moment at the column 
in an interior panel of the same dimensions. For this condition, the 
relation between the ratios of stiffness of the column and the slab 
should be as expressed in equations (3), p. 313. 

Where the wall columns are much stiffer, negative bending 
moment in the slab at the wall columns should be increased. 

Bending Moments in Exterior Columns.—The negative bending 
moment developed in the slab at the wall, the magnitude of which is 
discussed above, is transferred to the exterior columns and must 
be resisted by them. In the roof, the total bending moment is 
resisted by the column below. In lower stories, the bending moment 
is resisted partly by the column above and partly by the column 
below. The proportion of the moment resisted by the two columns 


depends upon their ratio of stiffness, £ The stiffer column will 
resist a proportionally larger share of the bending moment. 

The bending moment in each column may be found by dividing 
the bending moment in the slab by the sum of the ratios of stiffness 
(the moment of inertia [., divided by the length h, of the two columns), 
and multiplying the result by the ratio of stiffness of the column 
under consideration. For practical purposes, it is accurate enough 
to assume that 60 per cent of the bending moment will be resisted 
by the lower column and 40 per cent by the upper column. 

It should be noted that the bending moment in the upper end of 
the column below the slab is negative, i.e., it produces tensile stresses 


COLUMNS IN FLAT SLAB CONSTRUCTION 315 


on the outside face of the column. The bending moment in the 
lower end of the column above the slab is positive and produces 
tension on the inside face of the column. 

Points of Application of Maximum Moments in Columns.—The 
maximum negative bending moment in the column, as determined 
in the above paragraph, may be considered as acting at the bottom 
of the bracket, or, when no brackets are used, at the bottom of the 
slab or drop panel. The maximum positive bending moment acts 
above the slab. Plotting the maximum negative moment at the 
top and the maximum positive bending moment at the bottom, and 
connecting the points by a straight line, the moments at intermediate 
points may be determined. 

Steps in Design of Exterior Columns.— After the column load and 
bending moments are determined as described above, it is necessary to 
compute the maximum compression stresses for the combination’ of 
the concentric vertical column load and the bending moment, and to 
determine the amount of tensile reinforcement required to resist the 
tensile stresses for the most unfavorable combination of the con- 
centric column load and the bending moment. 

The first step in this process is to determine the dimensions for a 
column required by the concentric column load alone. This con- 
sists simply in applying the ordinary rules for column design, pp. 
403 to 469. 

Next, the maximum compression unit stresses in this column 
are computed for the combination of maximum concentric column 
load and the maximum bending moment, either by formulas on 
p. 175, or formulas on p. 182, depending upon the amount of 
tensile stresses developed for this condition of loading. The maxi- 
mum unit stresses thus obtained should not exceed the maximum 
allowable stresses as recommended on p. 463. In computing the 
combined stresses, the full section, including the fireproofing, may be 
used in the same manner as in slab at the support. If the computed 
unit stresses in concrete exceed the allowable unit stresses, the sec- 
tion should be increased. 

The final step is to determine the required amount of tension 
steel to resist the bending moment. For this purpose, that combina- 
tion of the vertical column load and the bending moment which 
produces maximum tension should be used. This combination will 
be different from that used for determining maximum compression. 

There is a difference of opinion as to the method of determining 


316 DESIGN OF FLAT SLAB STRUCTURES 


the required amount of reinforcement to resist the bending moment 
in the column. 

The following method, developed by the authors, is recommended 
for use. 

For the top columns in a structure, where the vertical column 
load is small in comparison with the bending moment, the amount of 
tensile steel should be determined for the bending alone, the effect 
of the vertical load being disregarded. The amount of steel is 


determined from the beam formula A, = ai where d is the distance 


from center of tension steel to the face of the column. If the amount 
of steel required by the column design is not sufficient, additional 
bars should be used. These do not need to extend the full length 
of the column, as the bending moment decreases rapidly. 

For columns in lower floors, where the total dead load carried by 
the column is appreciable, the following method for computing 
tensile stresses in column is recommended by the authors. 

Assume that the structure above the floor under consideration 
is loaded with dead load only, but that the floor under consideration 
is loaded by dead and live load. Under such conditions, provide 
sufficient tensile strength in the column to give a factor of safety of 
two in case of overloading of the floor (with the structure above not 
loaded). 

For this purpose, compute the dead load on the column above the 
foor under consideration. Add to it double the reaction due to live 
and dead load from the floor. 

Compute the bending moment in the column produced by the 
dead and live load on the floor under consideration. 

Multiply this moment by two and combine it with the column 
load computed above. Provide such an amount of steel in the column 
that the tensile stresses will not exceed double the allowable tensile 
stresses in steel. If the amount of reinforcement required by column 
design is not sufficient additional bars must be added. 

This method gives the same factor of safety, against cracks in the 
column, as is used in the floor construction. 

The method will be understood from the following explanation. 
It may easily happen during the life of a structure, and even during 
construction of the structure, that the wall panels of one floor are 
overloaded, while no appreciable live load is placed on the floors 
above. With such loading, only the dead load above this floor can 


COLUMNS IN FLAT SLAB CONSTRUCTION 317 


be counted upon as reducing the tensile stresses produced by the 
bending moment. This dead load is constant. It will not increase 
with the overloading of the slab. When the slab is overloaded by 
double the design load, the column load below the slab will be 
increased by the reaction of the slab, but the column load above the 
slab will not be changed. | | 

Some engineers and some building codes (notably that of the 
City of Chicago, Flat Slab Ruling) require that the amount of steel 
for the bending moment in the column be computed for the bending 
moment alone, and that the regular beam formulas and the same 
allowable tensile stress in steel be used here as for parts of the build- 
ing subjected to bending only. The amount of steel thus obtained 
is required in addition to the regular column steel. The steel may 
be arranged as shown in Fig. 109, p. 318. This method is too 
wasteful. 7 

Other specifications permit the maximum direct column loads to 
be combined with the bending moment, and if the resulting tensile 
stresses in steel do not exceed the allowable unit stresses, the column 
is considered satisfactory. This method is not safe. It depends 
upon the total dead and live load on the column to balance the tensile 
stresses, although it is clear that the total column load is not always 
there to balance these stresses. Also, in case of overloading of one 
floor, the bending moment in the column is increased but the col- 
umn load above the floor remains the same. The method already 
described, on the other hand, safely provides for such contingencies. 

Anchoring of Steel Resisting Bending in Wall Column.—The steel 
designed to resist bending stresses in the column should be properly 
anchored, else the bars may pull out instead of resisting tension. 

In the top of the top story columns, the outside bars should 
extend, above the points of maximum bending moment, a sufficient 
length to develop the strength of bar in tension. If this is not 
possible by straight imbedment, the bars should be properly hooked 
at the top in the slab. The inside bars are compression reinforce- 
ment. 

In intermediate stories, the outside bars at the top of any column 
are sufficiently developed when they are extended into the column 
above, the proper distance from points of maximum moment to 
develop the strength of the bar in tension. The inside bars of any 
column serve as compression reinforcement below the slab and as 
tension reinforcement above the slab. The length of imbedment of 


318 DESIGN OF FLAT SLAB STRUCTURES 








WL 
__--- Bend. Mom. Steel ~35- 


Roof Slab 


Regular Vertical 

and Spiral 

Reinforcement 
i 


h 
3 






SECTION A-A 


Not less than 


v Floor Slab 





eet oe Bend. Moment Steel ae 
ar wor 
1-2] 05 
| 
Ae 
ua 
A} 2 
hes 
Soi ™ 
Slow 
| i=) 
a= 
, 
' 
H 
' 





Frc. 109.—Wall Column Reinforcement, Chicago Ruling. (See p. 317.) 


COLUMN HEADS 319 


these bars in the column above, measured from the top of slab, 
should be such as to develop the full tensile strength in the bar. 

In all the above cases, an imbedment equal to 40 diameters of the 
bar for deformed bars, and 50 diameters for plain bars, may be con- 
sidered ample. | 


COLUMN HEADS 


The column head usually has the shape of a truncated cone or a 
pyramid. Its cross section depends upon the cross section of the 
column. For round columns, the cross section of the head is also 





Round Column Head Octagonal Column Head 
for Square Column 


Fig. 110.—Typical Design of Column Heads. (See p. 319.) 


round; for square columns, it is sometimes square. A more pleasing 
effect, however, is obtained by making the section octagonal, as 
shown in Fig. 110. Sometimes, when metal forms are available, 
the column heads for square columns are made circular in section. 
The effect is not pleasing, as the joint between the column and the 
head does not look neat. For octagonal columns, octagonal heads 
are used; for oblong columns, the heads may be as shown in Fig. 111. 
The forms for such heads may be made of sheet metal, the two 
halves of a circular column head being joined with two flat pieces of 
sheet metal. 


320 


The column heads may be made ornamental. 


DESIGN OF FLAT SLAB STRUCTURES 


Metal forms for 


the column head shown in Fig. 112 are readily obtainable. 





Fig. 111.—Column Head for Oblong Column. 
(See p. 319.) 


At wall columns, when 
spandrel beams are used, 
brackets are substituted for 
column heads, as shown in 
Fig. 113. Some codes, how- 
ever, require that a section 
of a regular column head be 
provided at the wall column. 

Designof Column 
Heads.—The function of 
the column head is two- 
fold: first, to decrease the 
shear; second, to reduce the 
net span and thereby the 
critical bending moments. 
It is a well-established rule 
that, in order to fulfill the 
second function, le., to 
decrease the span, the sides 
of the column head should 


form with the horizontal an angle of not less than 45 degrees. 
Also, the diameter of the column capital to be used in calculations 
should be taken where the vertical thickness of the column head is 


at least 14 in. The theoretical shape 
of the column head is shown in Fig. 
110. All rules are based on round 
column heads. When square column 
heads are used, they may be considered 
as equivalent to a round column head 
with a diameter 20 per cent larger than 
the side of the square. 

If it is desirable, for architectural 
reasons, to use for the column head 
a flatter angle than 45 degrees, as 


shown in Fig. 114, p. 321, the effective 





at mike Diameter---- 


Fig. 112.—Ornamental Column 


Head. (See p. 320.) 


column head is then governed by lines drawn at 45 degrees, as shown 


in the figure. 


If a block is placed above a column, as in Fig. 115, it may be con- 


COLUMN HEADS 321 


sidered as a column head and its diameter determined as shown in 
the figure. 





--Wall Column 






\~--. Bracket 


Section A-A Section B-B 
Fig. 113.—Wall Bracket. (See p. 320.) 


Early rules (some of which are still in use) specified a definite 
relation between the span of the panel and the diameter of the 





Fig. 114.—Flat Column Head. (See ps 320,) 


column head, without regard to the design load. In such specifi- 
cations, the bending moments were computed on the basis of the 
specified’ size of the column and were 
expressed as functions of the gross span 
and the load. Modern rules do not im- 
pose any arbitrary restrictions as to the 
minimum diameter of the column head. 
Its size is only limited by the shearing 
stresses in the slab. The selection of the 
proper size depends upon economy of 
design and also upon the architectural pig 115.—Block Serving as 
requirements. As the bending moments Column Head. (See p. 320.) 
are functions of the net span, the increase 

in the diameter of the-column head decreases the bending moments 
in the slab, and vice versa. 





322 DESIGN OF FLAT SLAB STRUCTURES 


The minimum diameter of the column heads specified by the 
Chicago and New York City regulations equals 0.2251 as given on 
page 397. The authors recommend that the size of the column 
head be varied with the design load. For factories and ware- 
houses, the diameter of the column head may be equal to 0.225 to 
().251, where J is the span in square panels and the average span in 
rectangular panels. For light manufacturing buildings, 0.22 may be 
used. 

In buildings carrying light loads, such as office buildings and 
apartment houses, the column head may be greatly reduced or even 
omitted if desirable for architectural reasons. 

For the method of computing shearing stresses at column head, 
see p. 346, under Shearing Stresses. 


DROP PANELS 


By the term drop panel, is understood a thickening of the slab 
around the column. ‘This portion is sometimes called the plinth, or 
simply the drop. The drep panels in square or nearly square panels 
are usually square. In rectangular panels, rectangular drops may 
be used, with a ratio of dimensions of drop equal to the ratio of the 
corresponding sides of the panel. Sometimes, for the sake of appear- 
ance, the drop panels are made round or octagonal. This compli- 
cates the formwork, and the effect does not warrant the departure 
from general practice. 

Advisability of the Use of the Drop Panels.—In early flat-slab 
designs, no drop panels were used. This gave unsatisfactory results 
inmany cases. The shearing and compression stresses at the column, 
in many designs, were too high and the deflection of the slab exces- 
sive. In recent years, the practice went to the other extreme of 
using drop panels indiscriminately even in cases where they could be 
omitted with economy. The designer should keep in mind that, 
from the standpoint of the user of the building, drop panels are not 
desirable. This is particularly the case where shafting is suspended 
from the under side of the slab. As a general proposition, then, drop 
panels should not be used except for economy, or when it is not 
possible to satisfy the stresses at the column without them. In 
deciding whether or not to use drop panels, comparative estimates 
of cost should be made of designs with and without drop panels, 
taking into account not only the cost of the materials but also the 
extra cost of formwork for the drop panels. 


DROP PANELS 020 


It will be found that for live loads over 150 Ib. per sq. ft., drop 
panels are economical. Tor smaller loads, down to 100 Ib. per sq. ft., 
drop panels are economical only for long spans and for specifications 
requiring bending moment coefficients in excess of those recommended 
by the authors (such as 1916 Joint Committee Recommendations or 
the Boston Code). In many cases, the use of larger column heads, 
with some compression reinforcement in the slab, may be found more 
economical than the drop panels. The cost of extra steel may be 
more than balanced by the saving in concrete and in the cost of 
formwork for the drop. For loads of 100 Ib. per sq. ft. and under, 
drop panels are hardly ever economical. 

Dimensions of Drop Panels.—The dimensions of the drop to be 
determined are the depth and the width. These are to some extent 
interdependent, as will be explained below. 

The purpose of drop panels is to reduce the shearing unit stresses, 
both at the column head and at the edge of the drop panel ; also to 
reduce the compression stresses and the amount of tensile steel at’ 
the column head. The magnitude of the shearing and compression — 
unit stresses at the column head depend upon the depth of the drop 
panel. The most economical depth is one for which the compression 
unit stresses in the concrete are about equal to the maximum allow- 
able unit stress. Additional depth saves steel but increases the 
amount of concrete in a larger ratio. Instead of adopting an arbi- 
trary depth for the drop panel, the designer is advised to compute it 
by formulas on p. 389. Since it is advisable to use the same depth 
for all drop panels throughout the floor, the depth should be com- 
puted for the largest bending moments. Ordinarily, these will 
occur in end panels, i.e., at the first interior columns next to the 
wall. ‘The compression stresses for interior panels will then be some- 
what smaller than the maximum allowable. 

The width of the drop panel depends upon three factors, the 
first being the compression stresses at points of maximum bending 
moment. As evident from the discussion on p. 341, under “ Com- 
pression Stresses in Concrete,” and from Formula (29), p. 344, a 
large part of the compression stresses is resisted by the drop panel. 
If the compression stresses in concrete are nearly equal to the maxi- 
mum unit stresses, the width and the depth of the drop are inter- 
dependent. ‘The width in this case cannot be reduced without over- 
stressing the concrete in compression. 

The second factor is the diagonal tension stresses at the edge of 


324 DESIGN OF FLAT SLAB STRUCTURES 


the drop, as figured by Formula (41), p. 349. If, for a certain width 
of the drop panel, the shearing stresses are excessive, the width of the 
drop panel must be increased. 

The third factor is the compression and tensile stresses at the 
edge of the drop panel, where the depth of the flat slab is suddenly 
reduced by the depth of the drop. This reduction is permissible 
only when the bending moment becomes so much smaller that, with 
the decreased depth, the compression and tensile stresses do not 
exceed the allowable unit stresses. The location of the section where 
the drop panels may be stopped will depend mainly upon the ratio 
of the depth of the drop panel to the total depth at the point of 
maximum bending moment. For larger ratios of the depth of the 
drop to the depth of slab, the required reduction in bending moments 
is larger than for smaller ratios. The points where the drop can be 
stopped are, therefore, farther from the edge of the column head for 
large ratios than for small ratios. Large ratios of the depth of the 
drop to depth of slab, therefore, require larger width of the drop 
than small ratios. : 

n various specifications, including that of the 1924 Joint Com- 
mittee, the width of the drop is arbitrarily fixed, irrespective of its 
depth, and varies from one-third to three-eighths of the span. This 
is not consistent. The authors recommend the following relation 
between the width of the drop panel and the ratio of the depth ‘of the 
drop pane! to the depth of the slab: 

Let d = effective depth of slab outside the drop panel; 

ty = depth of drop below slab; 
1 = span, center to center of columns, in the direction of the 
width under consideration. 

Then the width of the drop will be as civen in the table below. 


Ratio 2 Width of 
d Drop Panel 
0.1 0.281 
0.2 0.301 
0.4 0.331 
0.6 0.361 
0.75 0.381 


Drop Panels at the Wall Columns.—lIf drop panels are provided 
at the wall columns, they should be of the same depth as the interior 
drop panels. The width of the drop panel, parallel to the wall, 
should be the same as for interior drops in the same direction. The 


THICKNESS OF SLABS 325 


width at right angles to the wall, measured from the center of the 
column, should be equal to one-half the corresponding width of the 
interior drop panel. . 

In many instances, the shearing unit stresses at the wall columns 
are much smaller than at the interior column, on account of the 
larger circumference of the wall column, and also because part of the 
shear is resisted by the spandrel beam. Also, the bending moment is 
smaller. For these reasons, the drop panel may often be omitted 
at the wall column. The required amount of negative reinforcement 
at the wall must then be computed on the basis of the smaller depth. 
In rectangular panels, where the difference between the spans is 
appreciable, the drop panels at the wall in the short direction may be 
omitted. 


THICKNESS OF SLABS 


The thickness of the slab is governed, in flat slabs without drop 
panels, by the shearing and compression stresses at the column head. 
In flat slabs with drop panels, the only limiting stresses, as far as the 
thickness of the slab is concerned, are the diagonal tension stresses 
at the edge of the drop panel and the compression stresses in the 
center. The thickness of slab, determined by the stresses, may not 
be large enough to keep the deflection within desired limits. Since 
there are no reliable formulas for computing the deflection, most 
specifications impose a number of arbitrary and unnecessary limita- 
tions on the minimum thickness of the slab, such as that (1) the slab 
thickness shall not be less than 6 in., or that (2) it shall not be less 
than gz of the span for floor slabs nor less than zo of the span for 
roof slabs. 

It will be observed that the above limitations are general. They 
apply equally to flat slabs with and without drop panels. They apply 
also to all load conditions; the only distinction made is between 
floor slabs (irrespective of the design load) and roof slabs (also irrespect- 
ive of the design roof load). All these limitations originated when 
little was known about flat slab construction and are being carried 
along from specification to specification without investigation. 

The limitation of the thickness of slab to 6 in. originated when 
the four-way system was the only one in use and all flat slabs were 
built without drop panels. Under such conditions, it was not 
practicable to use thinner slabs and accommodate effectively the four 
layers of steel at the column. With the development of systems 


326 DESIGN OF FLAT SLAB STRUCTURES 


using only two layers of steel, and also with the introduction of drop 
panels, the reason for this limitation has been removed. With this 
limitation, flat slabs cannot be used economically for small spans. 
There is no reason why, for small spans, a thickness of slab propor- 
tional to the span should not be used. The authors recommend that 
the limitation be revised to permit 4-in. thickness for slabs with 
drop panels and 44 in. for slabs without drop panels. 

The second limitation of the thickness of slab to a certain ratio 
of the span was made for the purpose of limiting deflection. It was 
based on the performance, in use, of flat slab structures, mostly 
built without drop panels. Since, in a floor consisting of nearly 
equal panels and with slabs of the same thickness throughout, the 
wall panels deflect most, the limitation was based on the performance 
of the wall panels. Assuming that the limitation is correct under 
such circumstances, its use would be justified for buildings in which 
the interior and exterior spans are about equal and the thickness of 
slab is uniform throughout the floor, as then the conditions in the 
wall panel would govern. In cases where the wall panels are smaller 
than the interior panels, it would not be correct to apply this limita- 
tion to the larger interior panels. The unreasonableness of the rule 
‘s evident from the comparison of the conditions in Fig. 116, a and 6. 
Fig. 116a shows a building three panels wide, with exterior panels 
18 ft. and interior panels 26 ft. Fig. 1166 shows a building two 





Fic. 116.—Section through Flat Slab Buildings. (See p. 326.) 


panels wide, each of the panels being 26 ft. According to the limita- 
tion under discussion, the minimum allowable thickness of the slab 
is the same for both cases, although it is plain that the slab in the 
second case would deflect much more than in the first case. 

Another objection to this rule is that it separates flat slab design 
into floors and roofs, irrespective of the loading to be carried. In 
many cases this leads to absurdities. For instance; in an office 
building, the design live load for the floors is 70 lb. per sq. ft. 
Actually, this load is seldom reached. The load on the roof, on the 


THICKNESS OF SLABS 327 


other hand, may consist of cinder fill weighing about 60 Ib. per sq. ft., 
roofing weighing 10 lb. per sq. ft., and live load of 40 lb. per sq. ft. 
The total load, exclusive of the slab load, to be carried by the roof is 
110 lb. per sq. ft., of which 70 lb. is permanently on the roof. Accord- 
ing to the rule under discussion for 20-ft. panels, the required thick- 
ness of the floor slab would be 7} in. for the design load of 70 Ib.; 
while for roof slab carrying 110 Ib. per sq. ft——of which 70 lb. is 
permanently on the slab—only 6 in. would be used. 

Authors’ Rules Limiting Thickness of Slab.—It is questionable 
whether any limitation of the thickness of the slab, other than that 
imposed by the stresses, is necessary. General rules, however, if 
made consistent, may be of assistance to designers who are not 
grounded in flat slab design. Therefore, the authors recommend 
the following rules: 

Two groups are to be considered: (1) slabs carrying, in addition 
to the dead load of the floor slab, loads of 100 Ib. per sq. ft. and under; 
(2) slabs carrying, in addition to the dead load of the slabs, loading 
of over 100 lb. per sq. ft. 

The minimum thickness of the slab Se drop panels, for the 
first group, should be 70 of the end span or {7s of the interior span; 
for the second group, 35 of the end span or =; of the interior span. 
For slabs with drop panels, designed as Peiieried under proper 
heading, the minimum thickness given above may be reduced by 
3 of the thickness of the drop panel. 

Where the end spans are not monolithic with the wall columns, 
as in brick bearing jobs, the minimum thickness obtained above should 
be increased by 25 per cent. 

Thickness of Slab for Floors with Different Sizes of Panels.—If 
the panels are nearly uniform throughout the floor, the slab should 
be made of uniform thickness. The minimum thickness should then 
be based either on outside or inside span, whichever gives larger 
value. If the panels in one part of the floor are materially different 
from those in some other part, the thickness of the slab may be 
different for the different parts. The floor should be made level, 
and the difference in thickness made up by dropping the bottom 
surface of the slab. Where the different thicknesses of slab meet, 
the underside of the slab should be sloped gradually rather than 
stepped. 


328 DESIGN OF FLAT SLAB STRUCTURES 


BENDING MOMENTS IN FLAT SLABS 


General.—As explained in the section on theory, a flat slab is 
subjected to bending in two directions. For instance, at the column 
head and in the center of the panel, the slab bends in radial and cir- 








—e—e— 


ZA 


te 

te: reas 
~ Seat | 

Theoretical _7 


Cantilever 


rie 
oe | 
----> 
\ 
\ 
\ 


“eS e=—— 


/ 
Slab Suspended along Slab Suspended along 


Two Edges 


\ 










_Slab Suspended on 
~~~ Four Corners “~~~ 









bee @ mew oe oe we we wwe ew ew = = 


re ee ee eee see 


——< 
~— aaa swe sea anaes easererre= 







Suspended Portion 


Fic. 117.—Flat Slab Divided into Simple Parts. (See p. 328.) 


cumferential directions - while in the section between columns, the 
bending takes place in two directions at right angles to each other. 
The action will be better understood from Fig. 117, above, showing 


the flat slab divided into simple parts by cutting it along the points 
of inflection. 


BENDING MOMENTS IN FLAT SLABS 329 


To simplify the computations, it has become established practice 
to resolve the bending moments to which the slab is subjected into 
two sets of bending moments acting at right angles to each other. 
Each set produces bending in one direction only. The two sets, 
combined, produce the actual deflection of the slab. Each set of 
bending moments, and the required amount of reinforcement for 
each, can be computed separately. 

The character of bending due to each set of bending moments, 
when considered separately, is the same as if a continuous uniformly 
loaded slab were supported along two opposite panel edges. This 
condition produces negative bending moments at the support and 
positive bending moments in the central portion of the slab. 


1 - 1 Seation of Critical 
Neg. Bending Moment 


Column 
—<=>-- Strip 





eBending Moment 





Neg. Bending Moment 
tS) A 
- | 
Section of Max. Positive 


3,- 1,, Section of Critical 


ah 


ee, (ee ee ht (& \ ol PNT Cex 
aS, 31 Sr mey! 9h ee), (@) 


3,\ / 
ane ef coe Column 
‘ OE. = Head Strip 
3 - 3 Section of Critical 
Neg. Bending Moment 
Note: Solid lines indicate Critical Note: Solid lines indicate Strip in 
Section in one direction. Dash lines one direction. Dash lines indicate 
indicate Critical Section at right Strip at right angles. 
angles, : , 
Design Strip 


Sections of Critical Moments 


Fig. 118.—Sections of Critical Bending Moments and Design Strips. (See p. 329.) 


Sections of Maximum Bending Moments.—Sections of maximum 
bending moments are shown in Fig. 118, p. 329. As there are two 
sets of bending moments in a slab, there are also two sets of sections 
of maximum moments. Both sets are shown in Fig. 118. For the 
sake of clearness, Fig. 119 shows the sections of maximum moment 
for one set only. They are as follows: 

Section of Maximum Positive Bending Moment, coinciding with 
the center line of the panel. It is marked by 2-2 in Fig. 119, p. 330. 

Section of Critical Negative Bending Moment, coinciding with the 
edge of the panel and the edge of the column head. It is marked by 
7-1 in Fig. 119, p. 330. | 


330 DESIGN OF FLAT SLAB STRUCTURES 


Design Strips.—To simplify the computation of bending moments 
in any one direction, the panel is divided into strips running at right 
angles to the sections of maximum bending moments. (See Fig. 119.) 

The middle strip is concentric with the panel and is equal in width 
to one-half of the width of the panel. The column strips, placed one 
on each side of the middle strip, extend to the edge of the panel, so 
that the width of the two column strips is equal to the width of one 
middle strip. 

Similar strips can be drawn for the other set of bending moments. 


if 1-1= Section of Critical Negative 
rN Bending Moment 7 ee 


2-2=| Section of Max. Positive 
Bending Moment 


oe ess Se ee |--_---—---_—--—- 


i 





Bending Moments act at right angles to Sections 
l= Length of span in the direction of Bending Moments 
l= Length of span at right angles to | 


Note that the designation of | and l,depends upon the 
direction of Bending Moment and not upon relative 
length of spans, 


Fic. 119.—Sections of Critical Moments and Design Strips in One Direction 
Only. (See p. 380.) 


Distribution of Maximum Bending Moments.—The distribution 
of bending moments over the sections of maximum moments is not 
uniform. The intensity of bending moments is largest near the 
column and decreases towards the middle of the section. Instead 
of using the varying distribution of bending moments, it 1s accurate 
enough to assume a uniform distribution over the section in each 
strip, but to make the intensity in the middle strip smaller than in 


BENDING MOMENTS IN INTERIOR PANELS IN FLAT SLABS 331 


the column strip. The difference between the intensity of bending 
moment in the two strips is particularly large for negative bending 
moment. For the purpose of design, therefore, the total negative 
bending moment at each section is divided into two unequal parts. 
The larger part is assumed to act in the two-column head sections 
and the smaller in the middle section. A similar assumption is 
made regarding the distribution of the positive bending moment. 


BENDING MOMENTS IN INTERIOR PANELS IN FLAT SLABS 


The bending moments recommended by the authors, for use in 
computing design, agree substantially with those recommended by 
the 1924 Joint Committee. They are about two-thirds of the 
static bending moments. The reduction in bending moments is 
justified because the static bending moments do not take into account 
several factors which reduce tensile stresses in steel in flat slab con- 
struction. 

Let W = total live and dead load per panel, lb.; 

w = live and dead load, lb. per sq. ft.; 
1 = length of panel, center to center of columns, in the 
direction of computed bending moment, ft.; 
1, = width of panel, center to center of columns, ft.; 
c = diameter of effective column head, ft.; 
M = sum of positive and negative bending moments; 
M, = negative bending moment in two column strips; 
Mz = negative bending moment in middle strip; 
M3 = positive bending moment in two column strips; 
M4 = positive bending moment in middle strip. 


The formulas for bending moments given below apply to square 
and rectangular interior panels. For square panels, the bending 
moments in both directions are equal. For rectangular panels, the 
bending moments in the two directions are obtained by using, for / 
in the formula, the two dimensions of the panel, / and [;, successively. 

The sum of negative bending moment on section 1-1 and 
positive bending moment on section 2-2 is given by following formula. 
Sections are shown in Fig. 119, p. 330. 

Sum of Bending M ee, nterior Panel.— 


M = 0.09mi(1— 34) aed Lela toslegubesaanat (3) 


1.oswi(1 — 4) ries iad wile mete all (35 


ll 


oi 


302 DESIGN OF FLAT SLAB STRUCTURES 


Deve 27e\" 
Values of {1 — 3] ane he Us (lors 37) may be taken directly 


from table below. 


2c \2 2c \2 
Values of (1 ——-—) and 1.08 (1 —-— ,- 
31 31, 


a 


| C 
| Value of 1 


nese EEE 


0.1 |0.125| 0.15 |0.175| 0.20 | 0.225| 0.25 | 0.275 0,30 




















PAs . 
(: ie 0.870 | 0.841 | 0.810} 0.780 | 0.752 | 0.722 | 0.694 0.667 | 0.640 




















7) 2 
1.03(1-25) |0.940 0.908 | 0.875 | 0.842 | 0.812 | 0.780 | 0.749 Oaeh 0.691 


| 








The above sum on bending moments, M, may be distributed as 
given below: 


Negative Bending Moment on Section 1-1—I nterior Panels.— 


With Drop Panel Without Drop Panel 
Two Column Strips...... M, = — 0.54M M, = — 0.50M 
MiddlesStripayra te eee Mz = — 0.08M Mz = — 0.10M 
Positive Bending Moments Section 2-2—Interior Panels.— 
With Drop Panel Without Drop Panel 
Two Column Strips...... M3 = 0.23M M3 = 0.24M 
Middle: Stripsi.a de areas M, = 0.15M M, = 0.16M 


Permissible Variation in Distribution of Bending Moments.— 
A small variation in the distribution of bending moments, from that 
recommended above, is permissible. The positive bending moment 
in the column strips may be reduced by an amount not exceeding by 
0.02M, provided that the positive bending moment in the middle 
strip is increased by same amount. The negative bending moment 
in the two column strips may be either increased or reduced by an 
amount not exceeding 0.03M. If the negative moment is increased, 
the other three moments, namely the two positive moments and the 
negative moment in the middle strip, may be reduced, the sum of 
reduction not to exceed the increase. If the negative moment, is 
made smaller, the other three bending moments must be increased 
proportionally, the sum of increase to be at least equal to the reduc- 
tion of the negative moment, 


WALL PANELS 033 


Application of Formulas.—The formulas apply only to flat slabs 
extending in each direction at least over three panels, and only where 
the spans are equal or nearly equal. For bending moment formulas 
for unequal panels, as well as for flat slabs extending over two panels 
only, see Volume ITI. 

Use of Bending Moments.—The bending moments given above 
may be used ae for computing required areas of steel from the 
formula A, = —- or for computing stresses in steel from the formula 


a 
fae a where d is the effective depth of the construction at the 


section in consideration. Thus, in column strips, for negative 
bending moment the depth of the slab and drop panel, d1, should be 
used. For all other bending moments, the effective depth of slab 
alone should be used. 

For computing compressive stresses, the bending moments can- 
not be used directly, as it is necessary to make an allowance for 
factors which do not enter into the determination of the steel areas. 
For formulas for compression, see p. 341. 


WALL PANELS 


The design of wall panels is governed by the amount of restraint 
to which the slab is subjected at the wall. The restraint to be con- 
sidered is not only due to the wall column, but also to the wall beam, 
1.e., the beam at the edge of the wall panel. In practice, the follow- 
oe cases may occur. 

Case (1). Wall panels with properly designed concrete wall 
columns and with wall beams capable of resisting torsion. 

Case (2). Wall panels with properly designed concrete wall 
columns, but without wall beams capable of resisting torsion. 

Case (3). Wall panels supported on brick bearing walls. 

The authors’ recommendations for tending moments in the three 
cases are given below. They differ from the requirements of the 
1924 Joint Committee in that they differentiate between various condi- 


tions instead of making a general rule to cover all the dissimilar cases. 
Columns should have a ratio of rigidity of at least - = 1.5 4 
Let, in addition to the notation given on p. 331, 

3 = distance from center of column to outside of bracket at wall 


column. 


334 DESIGN OF FLAT SLAB STRUCTURES 


Then 


‘ 
we 0.09m71|1 us (3 i 7) ft.-Ib. 


af . eee 
lye Ci ; 
1.081] 1 _ a(; 4- 3) ins: 


W is the total load on wall panel, but does not include the spandrel 
load. lis the span length at right angles to the wall. It 3 is equal 


to _ values of M in the above formula equal the values of M for 
interior panel. 

The moments given below are fractions of the moment repre- 
sented by the above equation. 

Case (1)—Slab Restrained by Wall Columns and by Spandrel 
Beams.—This design is particularly recommended for flat slabs 
carrying heavy loads. The wall beams should be nearly square in 
cross section and capable of withstanding torsional moment as 
on p. 336. Moments are in terms of M, from equation (7) above. 


—With Drop Panel —Without Drop Panel— 


Ratio to Ratio to 
Moment TEOTnSH Moment pe 
in interior in interior 
panel panel 


Negative Moments at First 
Interior Column Line: } 
Column Strip, M,.... 0.62M eats 0.58M L156 
Middle Strip, Mz.... 0.12M 1.20 0.10M 1.20 


Negative Moments at, Wall 
Column 
Column Strip, M1.... 0.42M 0.80 0.40M 0.80 
Middle Strip, Me.... 0.08M 0.80 0.07M 0.80 


Positive Bending Moment. 
Column Strip, M3.... 0.27M 1d 0.28M 1.18 
Middle Strip, M,4.... 0.18M 1.20 0.19M 1320 


Case (2)—Slab Restrained by Wall Columns Only, Spandrel 
Beam not Capable of Resisting Torsion.—This occurs when no 
wall beam is used, or when the thin concrete spandrel wall above the 
slab is utilized as a wall beam. Such design should be used only for 
light live loads. The recommended bending moments are the same 
as for Case (1), with the exception of the bending moments in the 


WALL PANELS 


middle strip, both positive and negative, which are larger. 


in terms of M, from equation (7), p. 334. 


With Drop Panel 


Moment 


Negative Moments at First 

Interior Column Line: 
Column Strip, M@1.... 0.62M 
Middle Strip, M2.... 0.138M 


Negative Moments at Wall 
Column 
Column Strip, M@,.... 0.42M 
Middle Strip, Mz.... 0.08M 


Positive Bending Moment. 
Column Strip, M3.... 0.27M 
Middle Strip, M4.... 0.20M 


Ratio to 
moment 


in interior 


panel 


1.18 
1.30 


Moment 


0.58 
0.10M 


0.4012 
0.07M 


0.28M 
0.21M 


335 


They are 


—Without Drop Panel 


Ratio to 
moment 
in interior 


panel 


1.18 
1.30 


Case (3)—Slab Supported on Brick Bearing Wall.—No reliable 


restraint at the wall. 


—With Drop Panel 


Moment 


Negative Moments at First 

Interior Column Line: 
Column Strip, M,.... 0.75M 
Middle Strip, Me.... 0.11M 


Negative Moments at the 
Wall : 
Column Strip, M,.... 0.20M 
Middle Strip, Mo.... 0.08M 


Positive Bending Moment. 
Column Strip, M3.... 
Middle eee. 2 4 


0.32M 
0.21M 


Ratio to 

moment 

in interior 
panel 


1.40 
1.40 


1.40 
1.40 


Moment 


0.70M 
0.14 


0.20M 
0.08M 


0.34M 
0.22M 


The bending moments recommended, in 
terms of M, given by equation (7), p. 334, are: 


—Without Drop Panel 
Ratio to 
moment 
in interior 


panel 


1.40 
1.40 


1.40 
1.40 


The negative bending moment reinforcement at wall is required 
to take care of any restraint caused by the brickwork above slab. 

Permissible Variations in Bending Moment.—The variations in 
bending moment for the wall panels are the same as recommended 


for interior panels on p. 332. 


306 . DESIGN OF FLAT SLAB STRUCTURES 


Torsonial Moment in Spandrel Beam.—In Case (1) the spandrel 
beam should be able to resist a torsional moment at each end equal 
to one-half the moment M2 at the wall. For formula for torsional 
stresses see p. 95. 

Explanation of the Recommendations.—In wall panels, it has 
been observed that the loading produces not only appreciable bend- 
ing in the wall columns but also torsion or twisting in wall beams. 
The torsional resistance of the beams develops an equal amount of 
negative bending moment in the slab along the beam. This bending 
moment reduces the positive bending moment in the middle strip 
of the wall panels. If the wall beam is properly designed to resist 
the twisting moment, its resistance may be utilized by reducing the 
bending moments in the middle strip, as in Case (1). If no torsional 
resistance of wall beam exists, the bending moments in the middle 
strip must be increased, as in Case (2). 

In a number of cases, where flat slab has not been provided with 
proper spandrels, considerable trouble has been observed at the wall, 
where the brickwork above the slab or the thin concrete walls become 
distorted. Thin concrete spandrel beams have often bulged and 
cracked through torsion. 

Points of Inflection.—Points of inflection are points in the slab 
where the bending moments change from negative to positive. 
The bending moment at the points of inflection is, therefore, zero. 

For the purpose of determining the length of bars, the most 
unfavorable position of the points of inflection must be taken. 
For negative bending moment the most unfavorable condition 
occurs when all spans are loaded. For such case the distance of 
points of inflection from the edge of the column head may be assumed 
as equal to 0.211, where 1; = distance between the edges of column 
heads. ‘This location of points of inflection must be used in deter- 
mining the length of negative bending moment reinforcement. 

For positive bending moment the most unfavorable conditions 
occur when the adjoining spans are not loaded. For such condition 
the points of inflection are nearer the support and may be assumed 
equal to 0.1811. This location determines the length of positive 
moment reinforcement. 


FORMULAS FOR SLAB THICKNESS 


“The formulas for thickness of slab, given below, are not arbi- 
trary, but are based upon bending moments specified in previous 
paragraphs and upon allowable compressive stresses, as explained on 


FORMULAS FOR SLAB THICKNESS O37 


p. 341. These formulas and the formulas for compression given on 
p. 341, are based on the same general formulas, which in one case 
were solved for the depth of slab and in the other for the unit stress. 
If a depth of slab obtained from the formulas below is adopted, it is 
not necessary to compute the stresses. 

Formulas are given for thickness of slab at the column head and 
in the center of the column strip. If no drop is used, the required 
thickness at the column head governs the thickness of the slab. 
For slabs with drop panel, the thickness must be found both at the 
column and in the center. 

Let M, = negative bending moment in two column strips 

(see pp. 332 to 335); 
Mz = positive bending moment in two column strips 
(see pp. 332 to 335). 
t; = thickness of slab and drop at the column; 
thickness of slab in center of panel; 


1 = span length, center to center of columns, in direc- 
tion in which moments are considered; 
1; = span length, center to center of columns, at right 


angles to 1; 
c = diameter of column head; 
width of drop panel, in direction at right angles 
to span, /; 
j,k, = constants for specified stresses in concrete and 
steel f. and f; (see p. 205 and table, p. 880); 
w = uniformly distributed dead and live load per sq. ft.; 


o 
I 


C;, C2, C3, C4, Cs, Ce = constants for different ratios ; and different 


unit stresses (see diagrams, p. 911). 

Basis of Formulas.—The formulas for thickness of slab are based 
on the same principles as formulas for depth of slab in other concrete 
construction. The bending moments M, on pages 332 to 335 are 
recommended only for computing reinforcement. They do not 
take into account the increased compression stresses caused by 
arch action, the compression which balances the tension resisted 
by concrete nor the effect of the size of column head on compression 
stresses. To provide for these in the formulas given below the 
bending moments M, were multiplied by a constant equal to 
ae (1 — i) To get the thickness of slab, 1.5 inches were added 
to the effective depth. 


338 DESIGN OF FLAT SLAB STRUCTURES 


Thickness of Slab at Column Head as Determined by Negative 
Bending Moment.—The thickness of the slab at the columns, 
required by the compression stresses, should be computed from the 
following formulas. 


Flat Slab without Drop Panel. 
General Formula, as Determined by Negative Bending Moment, 


2c M 

t= 0.69(1 —=s ‘) sacl. 

37 SET e da ee 

Thickness ¢, will be in in. if M is in in-lb., f. in Ib. per sq. in., 

and I, in ft. Since, for interior panels, from formula on p. 332, 
2 

M, = 0.50M = 0.50| 1.08whi (1 — 34) | in.-lb., the formula 


for thickness becomes: 


Thickness of Slab for Interior Panels, 


0.51 207 a — 
where 
0.51 2c 
= 2280225). . 
il cea C 1 ( ) 
In exterior panels for slabs restrained at ends, the bending moment 
2 ~ 
is M, = 0.58 | 10st (1 — : ‘) in.-lb. For freely supported 


slabs, the fraction 0.58 changes to 0.70. The thickness of slab 
required for the two conditions is 


Thickness of Slab for Exterior Panels. Slab Restrained at End. 





0.56 2 6\ 77 = 
i= Sea — 35) Vy +1.5 = CalVw + 1.5 ROLL) 
where ; 
0.56 24 
— 1 St eae Be ° e e 7 ° * « e = » ° . 
e ae 3 4 | We) 


Thickness of Slab for Exterior Panels. Slab Freely Supported. 


Ben a | = 
t= $8139) Vw +1.5=C3livw+1.5 . (18) 





V fal seh 
or where " 
0.61 Ql 

c= SE (1-54) ee eee 


FORMULAS FOR SLAB THICKNESS 339 
Constants C1, C2, and C3 are given in diagram, p.911, for different 


ratios of ; and different unit stresses. 


The thickness, #, will be in in., if J is in ft. and w in lb. per sc. it. 


Flat Slab with Drop Panel. 
Thickness of slab and drop as determined by negative moment. 
General Formula for Thickness au Slab and Drop: 


. 2 c\ 
n = 0.50(1 ~ 39) 4/75 Th Sid baeaD Aint a tlh hi lt Soa Tay 
Thickness, ¢1, will be in in., if My is in in.-lb., b in ft., and f. in. lb. per 


Sq. in. 
Since, for interior panels, from formula on p. 332, 


2 
M, =0.54 M = 0.54] 1 .08whi?(1 ~ : ) | in.-lb., the formula for 


thickness becomes: 
Thickness of Slab and Drop for Interior Panels, 


per ee: 
ae et 31) w+ 1.5 = Calyfol + 1.5 (16) 





| Mien 8 t 
where ; 
0.41 PAT 
C4 ae Sg 3 1) ° . ° ° ° ° . ° ° ° ° . (17) 


in exterior panels, for slabs restrained at ends, the fraction 0.54 
in moment formula changes to 0.62, and for freely supported slab 
to0.75. The formula for thickness changes to: 

Thickness of Slab and Drop for Exterior Panels. End Slab 
Restrained. 


0.41 fut foe 
= Fel - 35) $1.5 =Cslyfut+1.5 (18) 


0.41 2 c\* 
Geet eee ibis erie JS geo ditasol 22 Ae eeh19 
; Jia 3 : ly) 


Thickness of Slab and Drop for Exterior Panels. End Slab Freely 
Supported. | 


Bree he ae 


0.45 2¢ 
=— ee Py e ° ° e ° ° ° e e ° AL 
Go eg 31) oe 


where 





where 





340 DESIGN OF FLAT SLAB STRUCTURES 


Constants C4, Cs, and Ce are given in diagram, p. 911, for different 
unit stresses and values of c. The thickness 1 is in in., if J, h, 0, 
and c are in feet and w in lb. per sq. it. 

Thickness of Slab as Determined by Positive Bending Moment. 
_The thickness of slab due to the positive bending moment should 
be based upon the largest value of the positive bending. This acts 
in the column strips and is given by formula for M3 on p. 334. The 
bending moments recommended on pp. 331 to 335 are for the purpose 
of computing the required amount of reinforcement. They do not 
‘nelude the bending moment resisted by tension in concrete. Since 
the thickness of slab, as governed by compression, must be based 
upon total bending moment, the value of M3 used in computing 
thickness of slab is multiplied by a ratio 1.5. This represents 
fairly well the ratio between the total compression and the amount 
of the tensile stresses carried by reinforcement. 


General Formula for Thickness of Slab. 


M3 
P= O.%nlaea tl... ee 
0 any nF (22) 
Thickness will be in in., if Mo is in in.-lb. and J, in ft. 


For interior panels, M3 = 0.24) 1. 080h? (1 — - ‘) | and the 


formula for thickness becomes: 


Thickness of Slab for Interior Panels, 





0.36 2c ad as ) 
where 
0.36 2c 
= ——— 1 rae a Na = 
& val 3 ) oe 


For exterior columns, the fraction 0.24 in equation for M3 may . 
be taken as 0.29 for slab restrained at end, and as 0.34 for freely 
supported slab. The formulas for thickness of slab become: 


Thickness of Slab for Extervor Panels. Slab Restrained at End. 





0.40 oe - 
i= deel 3 a ia ab ae 25) 
where : 
0.40 2¢ 
C — ff baer Ji ceeee . . . ° . ° ° . . . . 
‘ aa 3 ) ir 


COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 341 


In the above formula, it was assumed that the size of the bracket, 
c1, is equal to one-half the diameter of the interior column head. 
If no bracket is used, the depth of slab must be increased by a ratio 





Lug 

bes yBel 
of 

12° 

Soi 


Thickness of Slab for Exterior Panels. Slab Freely Supported at 


0.43 
A ee ee ee a Aan = SUA . i 
ae 3 51)! wtl=C/Vw+1 (27) 
where 
0.43 le 
C ——— 1 eee * ‘ . ' a . ‘ F : A 
faut 23) see 


Constants C7, Cs, and Cy are given in diagram, p. 911, for different 


Pet : : 
ratios 7 and different unit stresses. | 


~ Thickness ¢ is in in., if J is in ft. and w is in lb. per sq. ft. 


COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 


Not so much is known about the magnitude and distribution of 
compression stresses in flat slab construction as about tensile stresses 
in steel. The latter have been determined from extensometer read- 
ings in a comparatively large number of flat slab tests. Information 
on compression stresses from tests is much more meager. Since the 
modulus of elasticity of steel is known, it is possible in flat slab tests 
to get actual tensile stresses in steel by measuring deformations, and 
therefrom to get an idea of the magnitude of bending moment to be 
used in computing steel stresses. This, however, is not equally 
feasible in connection with compressive stresses, because the measured 
deformations are not a sure guide to the magnitude of the stresses 
in concrete. In the first place, the measured deformations of con- 
crete are not due to stresses alone, but consist of stress deformations 
and plastic deformations caused by a prolonged application of the 
stress. In the second place, the actual modulus of elasticity of the 
concrete, the correct value for which is necessary to convert deforma- 
tions into stresses, is usually unknown. While for steel the modulus 
of elasticity is practically constant, for concrete it varies not only 
for different concretes but also for different intensities of stresses in 
the same concrete. 


342 DESIGN OF FLAT SLAB STRUCTURES 


Compression Stresses in Slab at Column Head.—Tests show 
that in the column head section the stresses are not uniformly dis- 
tributed along the whole length of the section used for computing 
maximum bending moments. The stresses are much larger at and 
near the column head, especially at the diagonal lines where com- 
pression failures at high loads first occur. The size of the column 
head influences the magnitude and distribution of the compressive 
stresses. For equal bending moments and equal length of section, 
the stresses will be smaller for larger column head. 

Tests indicate that the sum of measured compression stresses on 
a section of maximum bending moment is larger than the sum of 
measured tensile stresses along the same section. Theoretically, the 
two sums should be equal as they are produced by the same bending 
moment. The difference is due to arch action, which increases 
compression stresses and decreases tensile stresses, and also to the 
fact that a certain percentage of tensile stresses is resisted by con- 
crete, so that the sum of tensile stresses based on steel alone is not 
the total sum of tension on the section.. ; 

The bending moments recommended on p. 332 are mainly for the 
purpose of determining tension reinforcement. They are based upon 
observations of tensile stresses in steel. Therefore, it is plain that 
it is not possible to use them for computing compression stresses 
without some modification. For this reason, in the formulas given 
below for determining compression stresses, the bending moments 
are multiplied by appropriate constants. 

In flat slabs with drop panels, there is another element of uncer- 
tainty as to the width of the slab to be used in figuring compression 
stresses. The drop panel and the adjoining slab deform as a unit 
and have a common neutral axis. The extreme fiber in the drop, 
being much farther below the neutral axis, will shorten more than the 
extreme fiber of the slab. The compression stresses, therefore, in 
the extreme fiber of the drop are larger than for the slab just adjoin- 
ing the edge of the drop. or points some distance from the edge of 
the drop panel, the deformation of the extreme fibers in slabs increases 
gradually until it reaches the same magnitude as the deformation of 
the extreme fibers in the drop. For instance, in the Channon test ! 
with 20 ft. 3 in. square panels, 6 ft. 6 in. square drop panels, 8 in. 


1 Test of a Flat Slab Floor of the New Channon Building, by H. F. Gonner- 
man and F. E. Richart: Proceedings of the American Concrete Institute, Sev- 
enth Convention. Vol. XVII, 1921, p. 182. 


COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 343 


slab thickness, 12 in. thickness at drop, for maximum test load the 
measured deformation of extreme fibers at the edge of the drop was 
0.00049, while the deformation of the extreme fibers of the adjoining 
slab was only 0.00020. At a distance of 6 in. from the drop, the 
deformation of the extreme fiber of the slab increased to 0.00033, 
and 14 in. from the drop, to 0.00053. This last deformation was 
even larger than the deformation of the drop. This indicates that 
while the effect of the slab at the column head next to the drop is 
small, it becomes considerable some distance from it. 

The difference between the stresses in the slab and in the drop 
will depend upon the ratio of the depth of the slab to the depth of 
the drop. For deep drops, the difference is greater and the distance 
is also greater from the edge of the drop to the point where the slab 
attains the same stresses as in the drop. 

With deep drops in the column head section, the effect of the slab 
outside the drop on compression stresses is negligible. In such cases, 
Formula (29), p. 349, may be used for computing compression 
stresses. This formula neglects the effect of the slab. 

With shallow drops, the effect of the slab outside of the drop may 
be considerable. In such cases, Formula (31), p. 344, will give 
better results. 


Formulas for Compression Stresses in Slab at Column Head.— 


Let M; = negative bending moment in column strips, in.-lb.; 
M, = tending moment resisted by concrete, in.-lb.; 
M, = bending moment resisted by compression reinforce- 
ment, in.-lb.; 
! = span of panel, center to center of columns, in direc- 
tion for which bending moment was figured, ft.; 
1; = span of panel, center to center of columns, perpendi- 
| cular to direction for which bending moment was 
figured, ft.; 
c = diameter of column head, ft.; 
b = width of drop panel measured at right angles to the 
span, ft.; 
d = effective depth of slab, in.; 
d, = effective depth of slab and drop, in.; 
A, = area of compression reinforcement, sq. in.; 
4, k, n = constants for concrete. See p. 207. 
Cio, C11 = constants. See Table 17, p. 912. 


344 DESIGN OF FLAT SLAB STRUCTURES 
Then 
Compression Stresses, Drop Only Considered Effective, 
C 
iz jkbd ° bd? 


0.25(1 =n ong 
gk 


lb. per Sq: hw euntae) 


where 


Cio = (30) 


In the above formula 6 is in ft. and dj in in. 


Values of Cio for different © and jk are given in Table 17, p. 912. 


l 
Compression Stresses, Drop and Slab Considered Effective, 


0. 25(1 2 1.25)\Mi 





lb. per sq.in.. (81) 


of 6D) Cefn) 


and Cio as given above. 
In the above formula J; is in ft. and d, in in. 


Values of C1; for different values of ; and a are given in diagram 


dy 
on p. 912. 
Compression Stresses, No Drop Panel Used, 


0.125(1 ca 1.25) Mi 
Je = jklyd? = C105 98 a2 





Ib=pereqeity a ae! 


Values of Cio for different ; and jk are given in diagram, p. 912. 


Lin ftsed7 anne 

In the above formulas, if My is in in. -lb., 1, 4, b, and c in ft. and 
d, in in., then f, is in lb. per sq. in. 

Values of jk.—In solving the equation for compression stresses, 
the values of jk must be assumed. For this purpose the values may 
be taken from table below. : 


Values of jk for Different Ratios of Steel 


p 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01 
jk 0.21 0.24 - 0.26, 0.28 0.80% 0:32 2g 0 ai Oise ates 


COMPRESSION STRESSES IN CONCRETE IN FLAT SLABS 345 


Use of Compression Reinforcement.—Compression reinforce~ 
ment may be found useful in many instances where it is not desirable 
to increase the thickness of the slab. Since, with equal spans, the 
compression stresses at the first interior row of columns next to the 
wall are larger than at other interior columns, it may be economical 
in flat slabs without drop panels to use compression reinforcement 
there, instead of increasing the thickness of the whole slab. The 
thickness of the slab can then be determined from the moment at the 
interior columns and made uniform for the whole floor. Also, in 
many cases, odd panels are much larger than the rest of the slab. 
It may be more economical to use compression reinforcement in 
such panels, instead of changing the concrete dimensions. 

Formulas for Compression Reinforcement at the Column Head.— 
Since the maximum compression stresses act at the column head, 
especially in the diagonal direction, the reinforcement, to be most 
effective, should be placed there. Using notation on p. 343. 

The bending moment resisted by the compression reinforcement 
(see p. 234) is 


M, = Al(n — =“aa ~ a). Ca Sieh Meas 


When the concrete dimensions are fixed and it is necessary to 
introduce compression reinforcement for the purpose of keeping the 
compression stresses within working limits, the problem may be 
solved as follows: 

Compute moment of resistance of slab at the column head, M,, 
based on the allowable working stress in concrete. To get as large a 
value as possible, the moment of resistance is based on the formula 
which takes into account not only the compression resisted by the 
drop panel, but also that resisted by the slab outside the drop panel. 
This moment of resistance may be obtained by solving Formula 
(31) for M,. Therefore, 

Moment of Resistance of Slab at the Column Head, 


Ci 
3 ed d 2 ce ° . e s e e es 4 
em i (34) 


The values of Cio and C1; are given on p. (912). 
If J; is in ft., d; in in., f. in lb. per sq. in., M- is in in.-lb. 
The bending moment to be resisted by steel equals the actual 


moment, M,, at the column head minus the moment of resistance, 
M,, computed above. 


M. = 


346 DESIGN OF FLAT SLAB STRUCTURES 


Thus, 
Moment to be Resisted by Compression Reinforcement, 
M, =M ~ M0 . .  e 
and 


Required Area of Compression Reinforcement, 
i ps k 
(n—Dfdd — a)k—-a 
If M, is in in.-lb., f, in lb. per sq. in., and d in in., A’, is in sq. in. 
Compression Stresses for Positive Moments. ere compressive 
stresses due either to positive bending moment or to negative bend- 
ing moment in the middle strip may be computed from the formula 
below. 
Let, in addition to notation on p. 343, 
M,, = bending moment in the section under consideration, in.-lb. 
Compressive Stresses for Positive Moment, 


ul 5M, 
Je gklid2 
Values of 1; are in ft. and those of d in in. 
Values of jk to be used in the above formula for different ratios of 
steel, p, may be taken from table on p. 344. 


tee 





(36) 


lb. per’sq: ine ae ee ae 


SHEARING STRESSES IN FLAT SLABS 


There is some difference of opinion as to the method of computing 
shearing stresses in flat slab construction and as to the allowable 
unit stresses at the various sections. The available test data on 
ultimate shearing resistance of flat slabs are meager. The tests on 
circumferential cantilevers and footings, which in some respects 
resemble the conditions in flat slab at the support, also give little 
information on diagonal tension. The test specimens either failed 
by tension (see p. 120) or the failure was complicated. 

The only reported tests on fat slabs carried to destruction are 
those conducted on the Western Newspaper Union Building ? during 
its demolition, and the tests at Purdue University on specially built 
panels designed according to the S. M. I. System and the Cor-plate 
System (see p. 99). None of these tests furnish any data on the 
ultimate resistance of concrete to diagonal tension, as all the slabs 

2Test of a Flat Slab Floor of the Western Newspaper Union Building, by 


Arthur N. Talbot and Harrison F, Gonnerman. University of Illinois Bulletin 
No. 106. 


SHEARING STRESSES IN FLAT SLABS 347 


failed by tension or tension in combination with compression. In 
the Purdue tests, cracks were observed at the column head, running 
diagonally downward from the tensile steel on the top of the slab 
toward the edge of the column head. The inclination of the cracks 
caused. them to be mistaken by some observers for diagonal tension 
cracks. Actually they were caused by the steel passing the elastic 
limit. The development of the tensile crack after the steel had 
passed the elastic limit was followed by a compression failure in 
concrete, a phenomenon observed in tensile failures of simple beams 
at points where there is no possibility of diagonal tension (see p. 22). 

It is‘generally agreed that the critical sections are at the column 
head and at the edge of the drop panel. Many codes and recom- 
mendations require that punching shear be figured at the edge of the 
column head and drop panel respectively. According to the Boston 
Code, the punching shear must be figured by the formula, v = ia’ 
and must not exceed 132 Ib. per sq. in. for 2 000-lb. concrete. The 
Chicago Code requires that the shear be figured by the same method, 
but the unit stress at the column is limited to 120 lb., and at the 
drop panel to 60 Ib. per sq. in. for 2 000-lb. concrete. The New 
York flat slab regulation specifies the same stresses as the Chicago 
Code, the difference being that the unit stresses are computed by 
dividing the total shear, V, by the area bd, instead of bjd. 

The modern tendency is to reduce the shearing unit stresses, 
especially at the column head. 

The authors recommend the following rule: 

Authors’ Recommendations for Shearing Stresses in Flat Slabs. 
—(1) Shearing unit stresses shall be computed at sections indicated 
in Fig. 120, p. 348. 

(a) At the column of critical shear the section is concentric with 
the column head and located from its edge a distance equal to 
Lio 14 in. 

(b) At the drop panel the section of critical shear is concentric 
with the drop panel and is located from its edge a distance equal 
to é— 1 in. } 

(2) Formula v = ia shall be used in computing the unit stresses 
where V is external shear in lb. at the section considered, 6 is the 
perimeter of the section in in., d, the effective depth in in. The 
resulting stress is in lb. per sq. in, 


348 DESIGN OF FLAT SLAB STRUCTURES 


(3) The shearing unit stress shall not exceed 0.03f'., which for 
1:2:4 conerete with f’, = 2.000 lb. per sq. in. equals 60 lb. per 
sq. In. 

This rule agrees substantially with 1924 Joint Committee Speci- 
fication, except that in formula for allowabie unit stresses the steel 
ratio, r, is omitted because there does not appear any justification 
for the same. 






Critical Section for 


Critical Section for nm 
Shear at Column--._. | 


Shear at Drop Panel 





BODE RD On 
SEN ir ikon Fhe a 


Fic. 120.—Section of Critical Shearing Stresses. (See p. 347.) 


Formulas and Constants for Computing Shearing Stresses.—The 
following formulas simplify the computation of shearing stresses. 
These may be used for square and rectangular panels, in. 

Let t; = thickness of slab and drop panel, in.; 

t = thickness of slab, in.; 
c = diameter of circular column head, ft. ; 
2(t1 as Lees) ft.: 








2. Radel 
b = side of square drop panel or long side of rectangular 
drop panel, ft.; 
> Otters Waco 
C3 0s Cre. Ae 
b, = short side of rectangular drop panel, ft. ; 
ba ipl le Le tre 
C4 > 91 12 ’ ap } 
1 = length of long side of panel, center to center of column, 
tz 


1, = length of short side of panel, center to center of column, 
tus; 


SHEARING STRESSES IN FLAT SLABS 


l 


ey ratio of short side to long side of panel; 


w = live and dead load per sq. ft.; 
v = shearing unit stress, lb. per sq. in.; 
d = effective depth, in. 


Then 
Shearing Unit Stress at the Column Head, 


C2 2 


a nt 101) 1) Derisqy 1M: 
3.147 X 12 3 
or 
y= Craw lb. per sq. in., 
where 
C2 2 C2 2 
m — 0.785() am — 0.785(5 
ome Le ee 
oa 12 B3102 
l l 
Note that / is in feet and d in inches. 
Shearing Unit Stress at Drop Panel, 
ee), | 
l 1 
vy = —————- 74 vl Ib. per sq. in., 
Ae x 12." 
or 
y= Cis, lb. per sq. in., 
where 
(i) 
ast 
Ci3 = ———_—.. 
42 


Note that I is in feet and d in inches. 


349 


(38) 


(39) 


(40) 


(41) 


(42) 


(43) 


Values of C2 and C13 are given in table on p. 913 for different 
values of co and c3, respectively, and different ratios of panel sides, m. 


For square panels, m = 1. 


390 DESIGN OF FLAT SLAB STRUCTURES 


When the drop panel is rectangular, the formula for shearing 
stresses changes to : 


y= Cis Ib. pet. sq. iD.,.. +. , ++» ytpip ene 
_where 
i + x “) 
Ci4 es = Sey aaa a aaa rer ee mere (45) 
21(% oe 1) | 


No table is provided for C14. 

Note that 1 is in feet and d in inches. 

Diagonal Tension Reinforcement in Flat Slabs.—The critical 
section, as far as diagonal tension is concerned, usually is at the column 
head. Generally, it is advisable to select such dimensions for the 
slab that the shearing unit stresses do not exceed the allowable 
stresses for concrete alone. If it is necessary to reduce the thickness 
below that required by shearing strength of concrete alone, stirrups 
or other diagonal tension reinforcement may be introduced, in the 
same manner as in any other type of concrete construction. 

The diagonal tension reinforcement should be computed to resist 
the excess of shear, beyond that which can be resisted by concrete 
alone. The method of procedure is as follows: 

The stirrups will be placed concentrically with the column head. 
Compute shearing unit stresses at the critical section. From these 
subtract the allowable shearing unit stresses for concrete alone. 
Multiply the difference by the perimeter of the section, in inches, at 
which shear was computed. Assume radial distance between two 
concentric rows of stirrups. Multiply the shear along the perimeter 
just found by this spacing. This gives total shear to be resisted by 
stirrups in the assumed spacing. Select diameter of stirrups and 
find the tensile strength of one leg. Divide the total shear to be 
resisted by the tensile strength of one leg of stirrups. This gives 
the number of legs of stirrups required to be placed around the 
column head. Next, compute shearing unit stress of a section 
at a distance from the critical section equal to the radial spacing of 
the stirrups. With this shear, proceed as before. 

The stirrups must be anchored to the tensile reinforcement, 
which in this section is at the top of the slab, and must extend of the 


bottom of the slab, where they should be provided with hooks at the 
ends. 


REINFORCEMENT FOR FLAT SLABS oo 


BOND STRESSES 


Bond stresses should be computed for the negative reinforcement 
at the column head by formula, 

Bond Stresses, 

= Sai (See p. 262.) 

They should not be larger than the allowable unit bond stresses, 
which for plain bars are equal to 0.04f’. and for deformed bars, 
0.05f'.. For 1:2:4 concrete the ultimate stress, f’, = 2.000 lb. 
per sq. in.; therefore, the allowable unit bond stress for plain bars 
is 80 lb. per sq. in. and for deformed bars 100 Ib. per sq. in. 

The following values should be used in the equation: 

W = total dead and live load on the panel; 
V = 0.35W for interior panel; 
0.40W for exterior panel; 
xo = perimeter of all effective bars in two column strips used to 
satisfy the bending moment, /;; 
d = effective depth at column head. 
The values 0.385W and 0.40W, respectively used for external shear 
V, are equivalent to the load producing the bending moment resisted 
by the reinforcement, the perimeter of which is used in the equation. 


REINFORCEMENT FOR FLAT SLABS 


Required Areas of Steel.—After the bending moments for the 
various strips are computed and the thickness of slab and drop panel 
selected, the required areas of steel are determined from the formula 


_M 
 jdfs 


7 


An approximate value of 7 = 7 may be used. For f; = 16000, the 
value of 7f; = 14 000, and for f; = 18 000, 7f; = 15 800 may be used. 

In the above formula, M is the bending moment for the various 
design strips and d the corresponding effective depth of slab from 
center of tension steel (which at column is in the top of slab) to surface 
of slab. In flat slabs with drop panels of the dimensions described 
on p. 323, the depth d at the column head equals the depth of the 
slab taken from the tension steel to the bottom surface of the drop, 
and the required area of negative bending moment reinforcement in 





(oe DESIGN OF FLAT SLAB STRUCTURES 


the column strips should be based on this depth. In computing the 
required area of steel at all other sections, the depth of slab without 
the drop should be used. When drop panels are omitted at wall 
columns, the reduced depth must be used in computing the required 
area of negative reinforcement in the column strips at the wall. 

To get the effective depth, Le., the distance from compression 
face of slab to center of tensile reinforcement, proper deduction should 
be made from the total depth. The distance to be deducted depends 
upon the number of layers of reinforcement. | 

The bars making up the effective reinforcement in any strip 
should be distributed over its whole width. They should be placed 
at right angles to the bending moment section and extended far 
enough on both sides, as explained in subsequent discussion. Con- 
versely, all bars cut at right angles by a bending moment section, 
if they extend far enough on both sides of it, as described under 
arrangements of bars, may be considered as effective at that section. 

The effect of bars in the horizontal plane, inclined at an angle 
other than right angle to a section of bending moment, as in the 
middle strip in a four-way system, may be considered as equal to the 
area of the bars multiplied by the sine of the angle of inclination. 
The same bars are also effective at a section at right angles to the 
above-mentioned section, and their effect is again equal to the area 
of bars times the sine of the angle of the bar with the second section. 
In square panels, the effect of diagonal bars is the same at both 
sections, while in rectangular panels the effect of diagonals is larger 
at the section parallel to the short side of the rectangle. 

Bars parallel to a section must not be considered as effective at 
that section. } 

The effectiveness of rings as used in the Smulski System, in 
resisting tensile stresses, 1s explained on p. 373. Obviously, the 
rules given above must be extended to fit the condition. 

Two-way System—In the two-way system, each design section 
cuts, in each design strip, only one band or group of bars running at 
right angles to the section. The area of the bars in each band should 
be made at least equal to the required area of steel in the strip in 
which the band is located. | 

In computing the effective areas for negative bending moment, 
not only the bent bars in the panel under consideration should be 
used, but also the bent bars extended from the adjoining panel, if the 
length of extension is sufficient to make them effective. 


REINFORCEMENT FOR FLAT SLABS BYa}s) 


Four-way System.—In the four-way system, the section of posi- 
tive moment in the column strips cuts the direct band of bars at 
right angles, and their area must be made at least equal to the required 
area of steel. In the middle strip, the section of positive bending 
moment cuts two diagonal bands at an angle. The effect of each 
band equals the area of steel in the band times the sine of the angle 
of inclination. Therefore, the area of bars in each band required 
to satisfy the bending moment may be obtained by dividing the 
total required area by twice the sine of the angle of inclination of the 
bars to the section. In a square panel, the angle of inclination is 
45 degrees and the sine is0.71. If A,4 is the required area of positive 
moment reinforcement in the middle strip, the area of bars in each 

Ass i) A 54 
) 7, STE AP Bs Ba 

In the column strips, the effective negative reinforcement con- 
sists of the bent bars in the band perpendicular to the section and of 
the vertical components of the bent bars in the diagonal bands. The 
components, again, equal the areas of the bars times the sine of angle 
of inclination. Not only the bent bars from the panel under consider- 
ation but also the bent bars extended from the adjoining span may 
be considered as effective, if the length of extension is sufficient for 
this. If the bars from the adjoining panel extend only far enough 
to develop their strength, they should not be counted as effective in 
this panel. 

Smuilski (S. M. I.) System.—In the Smulski System, the section 
of positive moment in the column strip cuts the trussed bars and (in 
square panels) all rings in Unit A and two or three outside rings in 
Unit B. 

The positive moment section in the middle strip cuts the remain- 
ing rings of Unit B and the diagonal bars. The effect of the diagonal 
bars is equal to their areas multiplied by the sine of their angle of 
inclination to the section. 

The negative moment section in the column strips cuts the rings 
of Unit C, the bent portion of the trussed bars of Units A and B, and 
the radial bars. The rings and the bars at right angles to the section 
are considered as fully effective, while the effect of the diagonal bars 
and radials is equal to their areas multiplied by the sine of their 
angle of inclination to the section. 

In all cases, the required area of reinforcement should be dis- 
tributed between the various units and the various parts of each unit. — 


diagonal band equals 


354 DESIGN OF FLAT SLAB STRUCTURES 


General Rules.—After the required amounts of steel at the sec- 
tions of maximum bending moments have been found from formula 
on p.. 351, the problem resolves itself into providing bars of such 
lengths, and bent in such a fashion, that not only the bending moments 
at critical sections but also those at intermediate sections are pro- 
vided for. 

In the region of the points of inflection, the positive and negative 
reinforcement should overlap, to take care of the movement of the 
point of inflection, due to partial loading. Tests show that, with 
columns rigid enough to withstand the bending moments due to 
unbalanced live load, the possible movement of the points of inflec- 
tion for different loading is small. 

If slender columns are used, the amount of positive moment 
reinforcement should not only be increased, but should also extend 
much nearer to the column than required with rigid columns. This 
means that, for this condition, the points of inflection for the maxi- 
mum positive bending moments are much nearer the column than 
for negative bending moment. 

Proper attention should be paid to bond stresses at the column 
head. There the external shear is large, and the rate of increase in 
tensile stresses is high, with the result that large bond stresses are 
produced. To keep the bond stresses within working limits, the use 
of bars of small diameter is advisable (see p. 262). 

Free ends of straight bars should extend beyond the points of | 
‘nflection for a length equal at least to 20 diameters of the bar but 
not less than 12 in. to insure the bond needed to make them available 
as tensile reinforcement at the points of inflection. 

It is important to remember that, until the tensile strength of bar 
is developed by bond, it is not the area of the bar that determines 
its capacity for resisting stresses, but the length of imbedment, - 
measured from the point where the stresses are investigated, to the 
end of the bar. For instance, in a {-in. plain bar, the full tensile 
working stress is not attained until the length of imbedment’® reaches 
8 ft.2in. Ina band of straight bars, therefore, contrary to common 
belief, the available area of steel is not constant throughout the bar. 
On the contrary, it is zero at the end and increases gradually as it is 
developed by bond. 


3 Where values governed by working stresses are given, normal concrete test- 
‘ing 2 000 Ib. per sq. in. at 28 days is assumed. 


REINFORCEMENT FOR FLAT SLABS 5915) 


When reinforcement consists of rings, it is not necessary to con- 
sider bond stresses except in determining the length of lap. 

When reinforcement consists of bands of bars, as in the two-way 
and four-way systems (see p. 358), most of the bars are of such length 
that they serve as positive and negative bending moment reinforce- 
ment. The arrangement of the reinforcement at the support is, 
therefore, dependent upon the reinforcement in the center of the span. 
In such cases, the number of bars required by the maximum positive 
bending moment is determined first; next, one of the schemes of 
bending, described below, is adapted. This fixes the number of bars 
which will be bent up and extended over the support. The area of 
the bent bars available for negative bending moment reinforcement 
is compared with the total required area of negative moment rein- 
forcement. If the bent bars are not sufficient, additional short bars 
at the support should be supplied. 

When reinforcement consists of rings and radials (see p. 367), 
the various units are designed more or less independently. 

Bending of Bars.—In band systems, a large proportion of the 
bars are bent up and extended over the support, as shown in Figs. 
122 and 127. The best method is to bend the bars before they are 
placed on the forms, as this insures proper location and proper height 
of the bends. Another method, which is much less satisfactory, is to 
place straight bars on the form and then bend them by means of a 
tool popularly called a ‘‘ hicky.” Bars of small diameter can be 
easily bent in this way; bars 3 in. and larger should not be hickied. 
This bending method is much less desirable than bending steel before 
placing. Satisfactory results with the “hicky ”’ are obtained only 
when thorough supervision is available. The length of time required 
to place the steel is increased by the time it takes to hicky the bars, 
and the placing of concrete is thus delayed. When the bars are bent 
on the ground ahead of time, this delay is obviated. The hickying 
of bars is particularly dangerous in rush jobs, where the concreting 
very closely follows the placing of steel. Under such conditions, it is 
obvious that exact bending of steel would be an accident rather than 
a certainty. 

In early flat slab designs, no actual bending of bars was done. 
The bars were supported above the form at the column and were 
allowed to sag down or drape. This practice was not satisfactory, 
as the sagging, especially with larger bars, was gradual, so that the 
bars did not get down to the bottom of the slab for quite a distance. 


356 DESIGN OF FLAT SLAB STRUCTURES 


Large sections, therefore, both at the top and at the bottom, were 
without effective reinforcement. As a result, cracks developed at the 
regions of sagging of the bars. Modern specifications prohibit 
draping of bars and require that bars be bent at proper definite places, 
and that both the positive and negative reinforcement be parallel 
to the surface of the concrete. 

1924 Joint Committee Rules for Arrangement of Reinforcement 
in Two-Way and Four-Way Systems.—The 1924 Joint Committee 
provides the following rules for arrangement of steel in two-way and 
four-way systems: 


“The design shall include adequate provision for securing the 
reinforcement in place so as to take not only the critical moments, 
but the moments at intermediate sections. Provision shall be made 
for possible shifting of the point of inflection, by carrying all bars in 
rectangular or diagonal directions, each side of a section of critical 
moment, either positive or negative, to points at least twenty (20) 
diameters beyond the point of inflection as specified in Section 151. 


This recommendation is rational. It provides general principles 
which will give satisfactory results. Following this, however, in the 
same section requirements are specified for the two-way and four- 
way systems which depart from the sound policy of laying down 
principles of design upon which the details could be developed to 
suit the conditions. By prescribing the lengths of bars, this rule 
practically limits the design of flat slab two-way and four-way rein- 
forcement to one arrangement. Several very satisfactory arrange- 
ments, both in two-way and four-way systems, described later, 
would be prohibited by this rule in spite of the fact that they are 
perfectly sound and give good results. | 

Support for Flat Slab Reinforcement.—Flat slab reinforcement 
should be securely tied and supported, in such a way that no dis- 
placement of the bars can take place during concreting. 

The displacement of bars most to be cuarded against is in the 
vertical direction, since the effect of both the positive and negative 
reinforcement depends upon its distance from the compression face 
of the slab. This distance must be as great as shown on the plans, 
while on the other hand, the bars must not be brought so near the 
tension surface of the concrete as to destroy the fireproofing. 

With positive reinforcement, the problem is to keep the bars high 
enough above the forms to get proper fireproofing. The best method 
is to support the bars on metal chairs of proper height. There are 


REINFORCEMENT FOR FLAT SLABS 007 


on the market, also, bar spacers which serve both as chairs and as 
spacers. Another method is to support the bars on small concrete 
blocks of proper height. To save the cost of such supports, in many 
cases, the bars are placed directly on the forms; then, during con- 
creting, they are lifted up by a special workman just after the con- 
crete is poured, and the concrete is allowed to flow under the bars. 
This method is not positive and should be prohibited, since the amount 
of concrete below the bars depends entirely upon the judgment and 
skill of the workman who lifts them. 

When pipes for electric wires are imbedded in concrete, they 
should be placed above and not below the reinforcement; otherwise, 
the reinforcement would be raised more than required by the plans. 
A crack is likely to form along the pipe. 

For negative reinforcement, supports are absolutely necessary. 
The bars must be kept the required distance above the form, since 
lowering of the bars reduces their effectiveness. In thin slabs, even 
small vertical misplacement may be dangerous. For instance, in 
an 8-in. slab without drop panel, with an effective depth of, say, 
6¢ in., an accidental lowering of bars by 1 in. would reduce the 
strength of the slab by about 15 per cent. 

In band systems, the negative reinforcement may be kept above 
the forms by four heavy supporting bars, resting either on concrete 
blocks of proper height or on metal chairs. The supporting bars are 
placed a short distance from the points where the bars begin to bend 
down. ‘The number of blocks should be sufficient to enable the sup- 
porting bars to keep the steel rigidly in place during construction. 

The negative reinforcement in the middle strip should also be 
kept a proper distance above the forms. For this purpose, at least 
two heavy supporting bars, running at right angles to the band, 
should be used. These supporting bars should rest on concrete 
blocks or chairs. In the four-way and circumferential systems, the 
negative reinforcement in the middle strip consists of short. bars. 
These are often placed after the concrete is in place but before it has 
attained its initial set. .This method requires proper supervision. 

In designing proper supports for the bars, it should be kept in 
mind that the reinforcement is subjected to very hard usage during 
concreting. Very often, even well-placed reinforcement, with sup- 
ports strong enough to keep it in place, is bent completely out of 
shape by the workmen, when they are spading and placing concrete. 
This is particularly true in cases where concrete is spouted. The 


358 DESIGN OF FLAT SLAB STRUCTURES 


adjusting of the spout requires a large number of men, who, occupied 
with their task, have no time to pay attention to the steel upon 
which they are walking. This emphasizes the absolute necessity of 
rigid arrangement of the negative moment reinforcement. 

Showing Reinforcement on Plans.—The best method of showing 
fat slab reinforcement on plans is to designate the various bands or 
units by letters, and give the reinforcement for each band or unit in 
a schedule. To explain the schedule, a typical exterior and interior 
panel should be prepared, clearly showing the arrangement of steel. 
The length and type of bending of various bars should be shown in 
sections. 

It is obviously insufficient to give the number of bars in a band 
required by the positive bending moment, without giving their 
length and clearly showing how the bars should be disposed, as then 
the arrangement of the steel at the support, which is the most heavily 
stressed section of the slab, is left to conjecture. This wrong method 
of showing flat slab reinforcement is unfortunately often found in 
general drawings submitted to the contractors for bids. No intelli- 
gent estimate of tonnage can be made on the basis of such plans. 
It leaves the field free to unsafe designs. 


FLAT SLAB SYSTEMS 


As far as arrangement of steel is concerned, flat slab construction 
may be divided into three eeneral groups, which, in the order of their 
appearance, are as follows: 


The Four-way System; 
The Two-way System; 
The Circumferential or Smulski System. 


The last-named system is also known under the abbreviated name 
of 8. M.1I. System. The three-way system may be considered as a 
modification of the four-way system. 


Four-WAyY SYSTEM 


The four-way system is distinguished by the use of four bands * 
of small bars at the column, running in four directions, namely, in 
two directions at right angles to each other—parallel to the center 
lines of the panel—and in two diagonal directions. While the 
Norcross patent was in force, the four-way system was exploited under 


4A band isa layer of definite width of bars spaced at definite distances apart. 


FLAT SLAB SYSTEMS 309 


several trade names. The best known were: Mushroom System, 
Cantilever Flat Slab System, Simplex System, and Barton Spider 
Web System. 

The Mushroom System was developed by C. A. P. Turner and 
used in a large number of buildings. At present it is used but little, 
if at all, because the features distinguishing it from ordinary four- 
way systems are either useless or harmful. The main reinforcement 
consisted of four bands of bars, running as in the modern four-way 
systems. At the columns, these bands rested near the top of the 
slab on a framework formed by “ elbow bars” and two rings placed 
on top of them. The “ elbow bars”’ were heavy bent bars with the 
vertical leg imbedded AA AAA 
in the column and the GFF RRRR or 
other leg extending into 
the slab, not horizon- 
tally, but diagonally 1 —— fais {BRO n aH 
downward. The band Cea o. 
of bars resting on the 
framework draped from 
the column. This left 
large sections of the 
slab, both at the top and LD, 
at the bottom, without aes: i 
effective reinforcement. 
The ‘ elbow bars ”’ and 
rings did not contrib- 
ute materially to the 
strength of the slab, Fic. 121—Barton Spider Web System. (See p. 359.) 
because, being located 
under the four bands of bars, they were too near the neutral axis. 

The Cantilever Flat Slab System and the Simplex System have 
practically the same arrangement of reinforcement. They differ 
only in the arrangement of chairs and spacing bars. 

The Barton Spider Web System has several patented features. 
The bottom reinforcement consists of four bands of straight short 
bars, which do not reach the column but extend on each side only a 
short distance beyond the points of inflection. The negative rein- 
forcement consists of two layers of bars running in two directions. 
This arrangement is shown in Fig. 121, p. 359, but is not recommended 
for use. 





360 DESIGN OF FLAT SLAB STRUCTURES 


Arrangement of Four-way System in Interior Panel.—Several 
methods of arranging flat slab reinforcement, now in common use, 
will be illustrated. Any of the first three methods may be used. 
The fourth method, although used extensively, is not recommended, 

First Method—Each band, rectangular or diagonal, consists of 
long bent bars and short straight bars. 

The long bars are bent at both ends near the points of inflection, 
and the bent portions are carried, near the top of the slab, across 
the column head beyond the points of inflection for negative reinforce- 
ment of the adjoining spans. These bars serve as positive bending 
moment reinforcement in one span, and as negative bending moment 
reinforcement in the same span and in two adjoining spans. 

The short straight bars, placed at the bottom of the slab, serve as 
positive bending moment reinforcement only. They should extend, 
on both sides, at least 20 bar diameters, but not less than 12 inches, 
beyond the points of inflection for positive moment reinforcement 
as specified on page 336. 

In this method, after the required areas of positive and negative 
reinforcement have been computed, the number of bars in the rect- 
angular and diagonal bands are determined from the areas required 
by the positive bending moments. A sufficient number of bars are 
bent up, both in the rectangular and diagonal bands, to supply the 
required area of reinforcement in the column head section. Not 
only the bent bars in the span under consideration, but also the bars 
extending from the adjoining spans, should be considered as effective 
‘n the column head section. The bars in each band that are not 
required for negative moment are carried straight. This method is 
illustrated in Fig. 122, folding page. | 

The negative reinforcement in the midsection consists of short 
straight bars. 

Second Method.—All bars in rectangular and diagonal bands are 
provided with one double bend, as shown in Fig. 123, folding page. 

They are of such length that one end extends at the bottom beyond 
the point of inflection and the other end is bent up to the top of the 
slab and extended across the column head, beyond the point of inflec- 
tion for negative reinforcement of the adjoining span. Each bar 
serves as positive reinforcement in one panel, and as negative rein- 
forcement in same panel and in the adjoining panel. The bars in . 
each band are so placed that a straight end alternates with a 
bent-up end. | . 









































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Notes: 


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is not identical. 

Distance of points of inflection from edge of column head equals: For negative moment 
reinforcement 0.2/;, and for positive moment reinforcement 0.181;, where 1; is the net span. 

Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but 
not less than 12 in. 

Area of straight bars at the bottom should equal at least one-quarter of the total area of 
reinforcement in column strips. 


Fig. 122.—Four-way Reinforcement. First method. (See p. 360.) 


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FLAT SLAB SYSTEMS d61 


The number of bars in each band is determined for the positive 
bending moment. In the column head section, every other bar of 
each rectangular and diagonal band is bent up aid becomes effective 
as negative bending moment reinforcement.’ Also, every other bar 
from the adjoining span is extended into this span and becomes 
effective as negative bending moment reinforcement. The area of 
the effective negative bending moment reinforcement is therefore 
equal to one-half the sum of the reinforcement at both positive sec- 
tions in the span under consideration, plus one-half of the correspond- 
ing sum in the adjoining span. With proper depth of drop panel, 
this amount of steel may be sufficient to supply the required negative 
moment reinforcement in the column strips. If not, additional 
short bars should be provided. This happens for shallow drops or 
for slabs without drop panels. The short bars should extend, on 
both sides, 20 bar diameters,® but not less than 12 in., beyond the 
points of inflection for positive moment reinforcement as specified 
on page 336. 

The negative reinforcement in the midsection consists of short 
straight bars. 

Third Method.—All bars in rectangular bands are bent up and 
extended, at both ends, 9 in. beyond the points of inflection for 
negative moment reinforcement of the adjoining panels. (See 
folding Fig. 124.) The bars in the diagonal bands are straight and 
extend at each end, 20 bar diameters beyond the points of inflection 
for positive moment reinforcement as specified on page 336. With 
such an arrangement, the negative bending moment reinforcement 
at the column consists of two layers only. The area of the negative 
reinforcement. is equal to twice the area of the rectangular band 
steel. If this is not sufficient to supply the negative bending moment 
reinforcement, additional straight bars should be added. 

_ The negative moment in the midsection consists of short straight 
bars. 

Fourth Method.—In this method, all four bands consist of long 
bars, all of which are bent up in the region of the points of inflection 
and extend beyond the column center. Some of the bars are extended 
into the adjoining panel, beyond the points of inflection, and are 
considered as negative reinforcement there. Other bars are lapped 
at the column. For comparatively short panels, some of the bars 
of the rectangular bands are made two panels long. 

’ See footnote, p. 354, 


362 DESIGN OF FLAT SLAB STRUCTURES 


This is the earliest method of arranging flat slab reinforcement 
and was used in the Cantilever and Simplex Systems. It is in dis- 
favor now, because all the bars are bent at the points of inflection, 
thus leaving a section of the slab, where positive moments may 
occur, without any positive reinforcement. This method is not 
recommended. 

Cross Bars-—In each of the four methods, negative bending 
moment in the middle strip (as defined on p. 330) is provided for by 
means of short bars placed across the rectangular bands. 

Arrangement of Four-way System in Exterior Panel.—In exterior 
panels, the continuous bands are arranged in the same manner as 
in interior panels. In discontinuous bands, the straight bars com- 
posing the band are extended to the wall column or wall beam. 
This applies also to the straight ends adjacent to the wall in the 
Second Method. The bent bars at the wall are provided with hooks 
of sufficient size to develop the strength of the bars. The hooked 
ends are extended to within 2 or 3 in. of the outside face of the 
column or wall beam. This last is an important requirement but is 
frequently neglected. 

Usually, it will be found that at the wall column the bent bars 
alone are not sufficient to resist the negative bending moment. The 
required area of steel should be computed, and extra, short, hooked 
bars should be provided. The hooks should be placed in the wall 
beam, as near its outside face as possible, and these bars should 
extend into the slab beyond the point of inflection. The required 
number of additional bars will be particularly large when the drop 
panel is omitted at the wall column, in which case the required area 
of steel must be computed on the basis of the depth of the slab and 
not the depth of slab and drop panel. 

In the midsection, at the wall beam, short bars should be used 
near the top of the slab. They should be placed at right angles to 
the wall beam. The hooks should be placed not more than two 
inches from the outside face of the wall beam. 


Two-wAY SYSTEM 


In the two-way system, the reinforcement consists of bands of 
small bars running in two directions. This system was originated 
and patented by Frank F. Sinks, and developed commercially by 
the Condron Company, under the trade name of the Acme System. 
The question of patents is not definitely settled. In one instance, 


FLAT SLAB SYSTEMS 363 


at least, the Sinks patent has been declared void by the court, but 
this is not necessarily a final decision. A typical arrangement of 
the Acme System is shown in Fig. 125, p. 363. 

Another variation of the two-way system is the so-called Cor- 
plate, developed by the Corrugated Bar Company. Some features 


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Fig. 125.—Acme System. (See D. 363.) 


of this arrangement are patented. This variation differs from ordi- 
nary two-way arrangements in that the steel in the column head 
section is not uniformly distributed through the band, but is more 
concentrated near the column than near the edges of the band. 
With bars of equal diameter, this is accomplished by making the 


364 DESIGN OF FLAT SLAB STRUCTURES 


spacing of the bars smaller in the middle of the band and then increas- 
ing it toward the edges of the band. This is shown in Fig. 35, p. 101. 

Arrangement of Two-way System in Interior Panel.—Several 
methods of arranging reinforcement are now in common use. Three 











“+ Points of 
// Inflection — 





“9" Hook A & BO ‘ A&B 


S90 Bar ‘~-20 Bar _ 90 Bar 
Diameters Diameters Diameters 
Bending Details 
Notes: 


Distance of points of inflection from edge of column head for positive and negative moment 


is not identical. 
Distance of points of inflection from edge of column head equals: For negative moment 
reinforcement 0.2i;, and for positive moment reinforcement 0.1811, where /; is the net span. 
Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but | 


not less than 12 in. 
Area of straight bars at the bottom should equal at least one-quarter of the total area of 


reinforcement in column strips. 


Fig. 126.—Two-way System. First Method. (See p. 364.) 


methods will be described and illustrated. Either of the first two 
methods may be used. The third method is described because it is 
sometimes used, but it is not recommended. 

First Method.—Each band consists of long bent bars and short 
straight bars. The long bars are bent up at both ends, and the bent 


FLAT SLAB SYSTEMS 365 


portions are carried, near the top of the slab, across the column into 
the adjoining spans, extending 9 in. beyond the points of inflection 
for negative moment reinforcement. The points at which the bars 
are bent should coincide with the points of inflection. 


|} Points of _ a __Points of _ 
"Inflection <- 7 Inflection ~ 
li, / 


Z 








\.9" Hook} <; | “x 
A,& Bf” Bending Details 


Section 


Notes: 

Distance of points of inflection from edge of column head for positive and negative moment 
is not identical. 

Distance of points of inflection from edge of column head equals: For negative moment 
reinforcement 0.2/;, and for positive moment reinforcement 0.18l;, where J; is the next span. 

Positive moment reinforcement to extend beyond points of inflection 20 bar diameters but 
not less than 12 in. 


Fig. 127.—Two-way System. Second Method. (See p. 366.) 


The short straight bars should extend, at each end, 20 bar diam- 
eters, but not less than 12 in., beyond the points of inflection for 


366 DESIGN OF FLAT SLAB STRUCTURES 


positive moment reinforcement as specified on page 336. This 
arrangement is shown in Fig. 126, p. 364. 

The number of bars in each band is determined from the positive 
bending moment. A sufficient number of bars should then be bent 
up to provide the required amount of steel for negative bending 
moment. With the arrangement recommended, the negative rein- 
forcement consists not only of the bent-up bars in the panel under 
consideration, but also of the bars extending from the adjoining 
span. It is good practice to have at least one-third of the bars in a 
band straight. If the negative moment reinforcement is not satisfied 
by the number of bent bars, additional straight bars should be used. 
These should extend, at each end, 9 in. beyond the points of inflec- 
tion for negative moment reinforcement. 

Second Method.—All bars in each band consist of bars straight at 
one end and are provided with a double bend at the other end. The 
straight portion serves as positive reinforcement and is of such 
length that it extends 20 bar diameters (minimum 12 in.) beyond the 
points of inflection for positive moment reinforcement. The point 
of bending of the bar coincides with the point of inflection. The 
bent-up portion of the bar is carried, near the top of the slab, into 
the adjoining panel, extending 9 in. beyond the points of inflection 
for negative moment reinforcement. Each bar serves as positive 
bending moment reinforcement at one section of maximum bending 
moment, and as negative reinforcement in the same panel and also 
in the adjoining panel. 

The bars in each band are so arranged that at each end the 
straight portions alternate with the bent-up portions. The total 
number of bars in each band is determined for the positive bending 
moment. The negative moment reinforcement will consist of one- 
half of the bars in a band in the panel under consideration, and one- 
half of the bars from the adjoining panel. Ordinarily, this is not 
sufficient in the column head section, and additional straight bars, 
extending, on each side, 9 in. beyond the points of inflection for 
negative moment reinforcement, should be used. This method is 
shown in Fig. 127, p. 365. 

Third Method.—In this method, all the bars in each band are 
bent up at both ends and carried near the top of the slab. Such an 
arrangement is not recommended, because it leaves a section of the 
slab, where positive bending moment may occur, without any rein- 
forcement. 


FLAT SLAB SYSTEMS 307 


Arrangement of Two-way System in Exterior Panel.—In exterior 
panels, the continuous bands are arranged in the same manner as in 
interior panels. In discontinuous bands, the straight bars or the 
straight ends of bars should extend to the column or wall beam. 
The bent bars at the wall column should be provided with hooks, to 
develop their full strength. These should be extended to within 
2 or 3 in. from the outside face of the column wall beam. This 
requirement is important. 

Usually, it will be found that, at the wall column, the bent bars 
alone are not sufficient to resist the negative bending moment. The 
required area of steel should be computed and extra, short, hooked’ 
bars should be provided. These should extend beyond the point 
of inflection. The required number of additional bars will be par- 
ticularly large when the drop panel is omitted at the wall column. 


CIRCUMFERENTIAL OR SMULSKI (S.M.I.) System © 


The Circumferential or 8.M.I. System was developed and 
patented by Edward Smulski. 

The reinforcement of typical interior and exterior panels is shown 
in Fig. 128, p. 369. A modification of the design is shown in Fig. 129, 
p. 370. . 

In general, the reinforcement consists of four types of units, 
which are usually designated by the letters A, B, C, and T. To 
distinguish between units of different panels, small letters are added. 
Thus, in end panels, the units become Ae, Be, Ce, Te, while at the 
wall Cw and Aw are used. When the steel in various floors is arranged 
differently, the floor is designated by a number placed before the 
name of the unit. Thus, 2A is Unit A in the second story, while RA 
is Unit A in the roof. 

Units A and B are placed near the bottom of the slab and serve 
as positive bending moment reinforcement, while Units C and T are 
near the top of the slab and serve as negative bending moment rein- 
forcement. . 

6 Because of the business association of one of the authors, Mr. Smulski, 
with the Smulski Flat Slab System, we have refrained from comparing in the 
test the relative economy of the different Flat Slab Systems. Personal analyses 
of the theory made by the signer of this footnote show that, using the same 
principles of design and same bending moment, the Smulski System requires 
20 to 24 per cent less reinforcement than the two-way or four-way systems. Tests 


of the system made by the signer of the footnote or witnessed by him prove the 
strength, reliability and permanence of the system.—Sanford E. Thompson. 


BOD. DESIGN OF FLAT SLAB STRUCTURES 


Bottom Reinforcement.—Units A are placed between columns. 
They consist of trussed bars and of rings. The number of rings 
depends upon the span and the load. For spans up to 22 ft., four 
rings are generally used. The ends of the rings are lapped to develop 
their full strength. The laps are staggered, so that not more than 
one ring is lapped in one section. 

The trussed bars are bent up at appropriate points and are car- 
ried, near the top of the slab, into the column head. In Fig. 128, 
they are extended into the adjoining panel and carried beyond the 
point of inflection. In the modification shown in Fig. 129, they are 
hooked on to the center ring within the column head. 

Units B are placed in the central part of the slab. They consist 
of rings and diagonal bars. The rings overlap those in Units A, 
thereby forming a continuous mat of steel extending over the whole 
slab. As evident from Figs. 128 and-129, the whole slab, with 
small exception of the area immediately around the column head 
which needs no such reinforcement, is provided with bottom rein- 
forcement capable of resisting stresses in any conceivable direction. 

The diagonal bars are bent up and carried, near top of the slab, 
into the column head, where they are hooked on the center ring. 

Units Aw, at the spandrel beam, consist of trussed bars and half 
rings that are provided with hooks at their ends. 7 

Top Reinforcement.— Units C are placed at the column, near the 
top of the slab. They consist of rings and radial bars. | 

One of the rings is placed within the column head, while the 
remaining rings are placed outside the column head. The number of 
rings depends upon the size of the panel and the load. They are 
spliced to develop the full strength of the bar, and the splices are 
staggered. The rings form the top layer of steel. 

The radial bars are in the shape of a hairpin, with prongs of uneven 
length. The longer prong is placed near the top of the slab, directly 
under the rings. The hook of the hairpin engages the center ring, 
which is placed within the column head. ‘The free end of the longer 
prong extends beyond the points of inflection. 

The reinforcement at the column head, in addition to Unit C, 
consists of trussed bars from Units A and B. ae 

Units 7 are placed across the panel edge in the midsection, con- 
sisting of bars of a length equal to one-half of the panel length. 

Units Cw, at the wall panel, consist of radial bars similar to those 
of Unit C, and of half rings placed on the top of the radial bars. 
The half rings are provided at their ends with hooks of proper 


FLAT SLAB SYSTEMS 369 


dimensions. To provide proper anchorage, the hooks are placed as 
near the outside edge of the wall beams as possible. 

Units Tw take the place of Units T at the wall. They are pro- 
_ vided on one end with a hook, which is placed as near the outside 
edge of the wall beam as possible. 


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\ a L. Column 


4 --<—__—— 
| Fradials Unit C,Cw 


Fic. 128.—Typical Smulski Flat Slab Reinforeement for Interior and Exterior 
Panel. (See p. 367.) 


EXPLANATION OF ACTION OF THE SMULSKI SysTEM 


Since the reinforcement of the Smulski (S. M. I.) System is 
radically different from that of the Two-way and Four-way Systems, 
its action will bé more fully explained below. 


370 DESIGN OF FLAT SLAB STRUCTURES 


The Smulski System takes advantage of the following prin- 
ciples: 

1. Ring reinforcement prevents any deformation of the concrete 
enclosed by it. Any tendency of the concrete within the ring to 
spread or elongate in any 
direction is resisted by 
the steel ring. There- 
fore, any tensile stresses 
within the ring, irrespec- 
tive of their direction, 
are resisted by the steel 
ring. 

A properly lapped ring 
is a complete unit and 
does not depend upon 
bond to develop stresses 
in the bar, as is the case 
with straight bars. While 
the ring is not at all effec- 
tive outside of its circum- 
ference, it is fully effec- 
tive at all points within 
: it. 

Fic. 129.—Modified Arrangement of Smulski Flat. Because of this prop- 
Slab Reinforcement. (See p. 367.) erty, a ring is a valuable 
reinforcement where the 
direction of stresses is either unknown or changeable, as In @ 
flat slab, particularly at the column head. In test, to destruction, 
of the Western Newspaper Union Building,’ with flat slab floor, 
four-way system, the stresses in diagonal direction were much 
larger than in rectangular directions, also at ultimate load the stresses 
‘n each band varied at an average from 18 000 lb. per sq. in. for out- 
side bars to 50 000 Ib. per sq. in. for center bars of the band. This 
means unequal utilization of steel, with the consequence that, to 
prevent excessive stresses in the bars carrying the largest stresses, 
a larger area of steel is required to resist the bending moment than 
if all bars were fully effective and equally stressed. 











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== Saintes =a see eeteenee ones ON he) 
‘Rings ‘Trussed Bars Lk / Radials 


Section Thru Column Head 


7 Test of a Flat Slab Floor of the Western Newspaper Union Building, by 
Arthur N. Talbot and Harrison F. Gonnerman. Bulletin No. 106, Engineering 
Experiment Station, University of Illinois. 


FLAT SLAB SYSTEMS oll 


With ring reinforcement, all the stresses are resisted by all 
the rings at their full working value, irrespective of their position 
and intensity. Thus in Purdue test, for maximum load, the stresses 
in rings at center column varied from 33 300 to 41 900 Ib. per sq. in. 
while in radials the variation of maximum stress was from 41 100 
to 44 700 Ib. (See p. 107.) 

Separating Flat Slab into Simple Parts.—It is a well-known fact 
that a continuous beam or slab can be replaced by simple parts 
without changing the stress conditions. Thus, the beam can be 
cut at the points of inflection, where the bending moment is zero. 
It is then changed into two cantilevers and a simple beam suspended 
from the cantilevers. The stresses and the bending moments would 
not be affected by this change. 

The same principle can be applied to flat slabs, which can be 
separated into following parts, as shown in Fig. 117, p. 328: 

Circular cantilevers at the column head; 

Slabs between columns; 

Slabs supported at four points subjected to stresses in all direc- 
tions. 

In designing the reinforcement, it is permissible to treat the 
separate parts independently and to provide in each of them a suffi- 
cient amount of steel to resist the particular moments to which they 
may be subjected. 

As the unit shear at the points of inflection is always low, not 
exceeding 40 lb. per sq. in., concrete is capable of taking care of the 
shearing stresses. 

Since it is not advisable to rely on concrete alone, the parts of the 
slab subjected to positive bending moment, and reinforced by Units 
A and B, are tied securely to the circular cantilever at the column 
head by the bent portions of the trussed bars and by the overlapping 
of Units A and C. 

The position of the points of inflection is variable for different 
positions of the live load. To provide for this and also to pre- 
vent secondary cracks due to temperature and shrinkage, the 
various units overlap, thereby tying the slab together and en- 
abling it to act as a whole if such action is required by any con- 
tingencies. 

Column Head Section.—At the column head the portion of the 
slab within the points of inflection acts like a circular cantilever, 
loaded uniformly over its area, and also along its circumference, by 


312 DESIGN OF FLAT SLAB STRUCTURES 


the loads transferred to it from the rest of the slab. This portion 
is subjected to negative bending moments; i.e., the particles in the 
upper part of the slab elongate, while the particles in the lower part 
are compressed. After deflection, the cantilever assumes the shape 
of an umbrella. 

The negative bending moment at the column head is larger than 
the positive bending moment in the center of the slab. The amount 
of steel required there is therefore larger than in any other part of 
the slab. To reduce the amount of reinforcing steel required at the 
column head, the depth of the slab in this section is often increased 
by forming the so-called drop panel. 

The most unfavorable condition of loading for the column head 
section is that which occurs when all the spans surrounding the col- 
umn are loaded. In such a case, the shape of the cantilever will be 
as shown in Fig. 130, p. 378. Since, after deflection, any circle 
increases its radius as well as its circumference, the particles must 
elongate in radial as well as in circumferential direction and are 
therefore subjected to radial and circumferential stresses. The 
most effective tensile reinforcement consists of rings and radial 
bars. 

Compressive stresses act also in radial and circumferential 
directions. The compression acting radially composes the bulk of 
compressive stresses. 

Slabs between Columns.—The principal stresses in this part act 
mainly in one direction, which at first is parallel to the edge of the 
sanel and then gradually becomes inclined. - In addition, negative 

stresses due to cross bending, and also due to shrinkage and tem- 
perature changes, act across the principal positive stresses. 

The advantages of using rings in this part to resist the various 
stresses are as follows: (1) They intersect the lines of equal defiec- 
tion more nearly at right angles than straight reinforcement. 
(2) They bind the Units A and B, thereby preventing secondary 
cracks. (3) The rings in the two units supplement each other. 
(4) The arrangement 1s economical, as the rings cover the whole 
surface without waste of material. 

Central Part of Slab.—The central portion acts like a slab sup- 
ported at four corners and loaded with uniform load. The bending 
moment is positive, so that the top is in compression and the bottom 
in tension. The stresses act in all directions; the reinforcement 
consisting of rings, therefore, is fully effective. 


FLAT SLAB SYSTEMS 373 


Action of Rings.—The rings resist all tensile stresses within them, 
irrespective of the direction in which they act. 

To understand the action of the rings, it is necessary to keep in 
mind that tensile stresses can not act without producing a corre- 
sponding lengthening of the materials. Also, it must be remembered 
that the rings are filled solidly with concrete. If the shape of 


ab&cd Arcs of Steel Ring before Deformation 
a,b, & c,d, Elongated arcs ab & cd after Deformation 


Points a oe oe eo Line of Inflection 


—_—— — __ 





son 
Lr Steel Ring 
“\\ also Points 
4-3 









Points 2a a roe 
a 





also Points 2- ]------ 






=-§teel Ring 
also Points 1 - 2-7 











PSteel Ring 
also Points - 










a 
ae 


*—Points 4° 







ne aes eo 


= — 


‘Neutral’ 
Axis 











4-4'=Elongation of distance 0-4 Section A-A 
SectionA-A  2-2'= Elongation of distance 0-2 

(equals 4- 4' plus Elongation 

of distance 2-4) 


Rings in Unit. C Rings in Unit B 


Fria. 130.—Action of Rings in Resisting Stresses. (See p. 373.) 


the concrete within the ring undergoes any changes due to tensile 
stresses, the tight-fitting rings must undergo ‘similar changes. 
When the slab is loaded and the concrete is elongated by stresses 
produced by the bending moments, it presses against the circumfer- 
ence of the enclosing ring and forces it to stretch proportionately to 
the magnitude of the stresses. 

Parts of the slab at the column and also in the center of the slab, 
subjected to stresses in all directions, are shown in Fig. 130, p. 373. 
AA is asection in any direction. When loaded, the slab is compressed 
near one surface and elongated near the other surface, so that in 


374 DESIGN OF FLAT SLAB STRUCTURES 


Fig. 130 the distance 0-3 shortens by 3-3 ft, while 0-4 expands by 
4-4 ft. The same is true of a section in any direction; therefore the 
circle 4-4 tends to assume the position 44 ft. Before assuming the 
new position, the concrete must stretch the ring by which it is enclosed 
and increase its radius and therefore its circumference. The concrete 
exerts a pressure along the circumference of the ring similar to the 
pressure of water In a reservoir. The steel ring, by its tensile resist- 
ance, partially prevents the movement, but it stretches to some extent, 
causing tensile stresses in the steel. 

Considering the second ring, it is evident that the movement of 
point 2 consists not only of the elongation of the distance 0-4 but 
also of the elongation of the distance 2-4. The outside ring, there- 
fore, shares the stresses with the inside ring. Any deformation of the 
concrete, irrespective of its direction, is taken up at once by all the 
rings placed outside of the place of deformation. All rings are 
effective in resisting stresses. . 

Forces Acting in All Directions.—Where the forces act in all 
directions, as in the center of a slab and at the column head, the ring 
stretches uniformly along its circumference. After deformation, the 
shape of the ring remains substantially circular and the stresses are 
uniform along its circumference. 

Forces Acting Principally in One Direction.—When the forces act 
principally in one direction, as in Unit A, the condition is similar to 
that of a solid disc of concrete with a tight-fitting steel ring around it, 
subjected to a force in one direction. Under the pressure of the 
enclosed concrete, the shape of the ring changes gradually into an 
oblong curve, with the concrete following and still pressing tightly 
on the ring. In this case the stresses in the ring are a maximum at 
the sections cut by a diameter perpendicular to the direction of the 
stress, and decrease to zero at points 90 degrees from the point of 
maximum stress. 

From the above, it is evident that the stresses due to the prin- 
cipal bending moment are small in the parts of rings of Unit A which 
are near the column head, so that they can resist stresses in diagonal 
direction in places where they run almost parallel to the diagonal 


bars. 
THREE-WAY SYSTEM 


The three-way system was developed by D. W. Morrow of 
Cleveland. It can be used only with a peculiar spacing of columns. 


OPENINGS IN FLAT SLAB BYES) 


The reinforcement then consists of three bands of small bars. The 
number of bars in each band may be determined for positivé bending 
moment. The arrangement of bars may be similar to that described 
in connection with the four-way system. The negative moment 
in the midsection should be provided for in the same manner as in 
the four-way system, and should consist of short straight bars. 


In Four -way System add 

at Column Diagonal bars 

to compensate for Diagonal-~_ 
Band omitted in this Panel. 


to carry Skylight 


Use Sufficient Amount of 
Reinforcement around 


opening 


ac) 
ied 
S 
= 
= 
= 
= 
D 
~ 
a) 
> 
a) 
> 
is 
= 
, 
8 
~~ 
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compensate for omission of 


Strip add Short Bars on top to 
Middle Strip Bands. 


‘2-2'Ra. 4 Ft. 0" Top 
—— at each corner 





‘2 "Rd. 72" on centers 


Section Through Skylight 


Fra. 181.—Opening for Skylight. (See p. 376.) 


OPENINGS IN FLAT SLAB 


Proper design of the flat slab at openings requires considerable 
judgment. Openings up to one foot in diameter seldom require 
framing unless they occur at the columns. For large openings, it is 
often possible to strengthen the slab sufficiently to get along without 


376 DESIGN OF FLAT SLAB STRUCTURES 


any beams. No general rule can be made to cover all cases. Inex- 
perienced designers are advised to provide beams at openings. 
Figure 131 shows a typical opening in the roof slab for skylight. 
With such an opening, the slab can be designed without any beams, 
the reinforcement on all four sides of the opening being made strong 
enough to carry the weight of the skylight. Sometimes, a curb is 


coe ees ee mee ee ee we a ae ew ws i ee ee 


Se A PS tet ee ee 





Fic. 132.—Opening for Staircase. (See p. 376.) 


provided, which may be built with the slab or separately after the 
slab is poured. If the curb is to be built separately, dowels should be 
provided in the slab, and when the curb is being built, the joint 
between old and new concrete should be bonded with neat cement 
paste to avoid leakage. ‘T’o prevent corner cracks, two -in. round 
bars should be placed diagonally at the top of the slab at each corner 
of the opening. 

Figure 132 represents another common type of opening, for a 


DESIGN OF BEAMS IN FLAT SLAB CONSTRUCTION BYi| 


staircase. A framing of beams is usually provided for such an 
opening. 


DESIGN OF BEAMS IN FLAT SLAB CONSTRUCTION 


When Beams are Necessary in Flat Slab Construction.—Some- 
times it is necessary to introduce interior beams in flat slab construc- 
tion, to carry for instance, heavy concentrated loads, or to frame an 
open‘ng. An illustration of the use of beams is given in Fig. 132, 
p. 376. 

Load Carried by Beam.—An interior beam in flat slab construc- 
tion should be designed for the load coming directly upon it, plus a 
flat slab load from a width of slab on each side of the beam equal! to 
one-fifth of the span at right angles to the beam. 

Often, in such cases, the beam is designed only for the load super- 
imposed upon it, and the flat slab, reinforced as if no beam were used, 
is assumed to carry the slab load. This is obviously wrong. The 
beam and slab cannot act separately, the behavior of the slab. being 
dependent upon that of the beam. Part of the slab on both sides of 
the beam acts as a T-flange for the beam, and, when the beam deflects, 
the slab deflects not as an independent slab, but as a flange of the 
T-beam. The position of the neutral axis is governed by the beam. 
The under side of the slab is in compression or, if not, the tensile 
stresses there are small. The effect of the slab reinforcement parallel 
to the beam (which would have been sufficient to carry the slab load 
if the beam were not there) is small; therefore, the load which would 
have been carried by the slab must be carried by the beam. Short 
bars must be placed at the top of the slab across the beam. 

Restraint of Interior Beams Framing into Columns.—Beams in 
flat slab construction are usually one panel long, and the tendency 
is to treat them as simply supported beams. This is not correct. 
The beams form a monolithic part of the flat slab construction. If 
they frame into the columns, they are restrained there to the same 
degree as the rest of the flat slab. When they frame into interior 
columns, they may be considered as fixed at both ends. When they 
frame into one interior and one exterior column, they may be con- 
sidered as fixed at the interior column and restrained at the wall 
column. i 

In Fig. 132, p. 376, for instance, beam B, is only one span long. 
If the rules for one-span beams were followed blindly, it would be 


378 DESIGN OF FLAT SLAB STRUCTURES 


ee Peis at 
designed for oo in the center, with little or no provision for bending 


moment at the supports. This would be not only wasteful but also 
unsafe. From inspection of the construction, it is evident that the 
beam is restrained not only by the column but also by the flat slab 
on the other side of the column. The beam has a restraining effect 
on the slab and in turn, the slab restrains the beam. 

Bending Moments in Interior Beams Framing into Columns.— 
Interior beams framing into columns should be designed as follows: 

The span of the beam should be taken as the distance between 
the faces of the column and not between the edges of the column 
heads. — 

The following bending moments should be used. 

Let J; = net span, i.e., distance between faces of columns, ft.; 

w = total live and dead load, lb. per lin. ft. 
Then 
Interior Panels.— 


Negative Bending Moment at Columns, 

M = — +,wli? ft.-lb. or — wh? in-lb. . . . . (46) 
Positive Bending Moment in the Center, 

M = + sul)? ft-lb. or 0.75wl)? n-lb. . = . (47) 


Exterior Panels.— 
Negative Bending Moment at Intervor Column, 


M = — pywl:? ft-lb. or — 1.2wh? in-lb. . . - (48) 
Negative Bending Moment at Wall Column, 

M = — +,ul,? ft.-lb. or — 0.75wh;? in-lb. . . (49) 
Positive Bending Moment in the Center, 

M = + pls? ft-Tb. or wh? in-lb. - 2) a (a0) 


Interior Beams not Framing into Columns.—Interior beams that 
do not frame into columns such as beam Bg in Fig. 132, p. 376, are 
only partly restrained at the ends. The bending moments should be 


2 


wl 
assumed as 
10 


be taken as from center to center of the supporting beams. The 
negative reinforcement should not be hooked in the beam but should 
be extended into the slab. 


2 
in the center and - at the support. The span should 


WALL BEAMS 379 


WALL BEAMS 


Wall beams are beams between exterior columns, used to carry 
the spandrels (i.e., the walls under the windows) or curtain walls. 

In flat slab construction, wall beams may be divided into two 
general types: (1) wall beams extending below the slab, and (2) wall 
beams extending above the slab. 

The various designs of the wall beams are shown in Figs. 133 to 
137. 





Fic. 133.—Wall Beams below Slab. (See p. 379.) 


Wall Beams Below the Slab.—When wall beams extend below 
the flat slab, as in Fig. 133, they must be designed to carry not 
only the wall load but also a portion of the flat slab load. The reason 
for this is explained on p. 388, and is evident from Fig. 134, p. 381. 
The best practice, endorsed by the authors, requires that 20 per cent 
of the slab load in the wall panel be considered as carried by the wall 
beam. The slab load should be assumed as uniformly distributed 
over the whole length of the beam. For example, if the span of the 
slab at right angles to the beam is 20 ft., the dead and live load on a 
width of slab equal to 4 ft. should be considered as being carried by 
the beam. 

The authors recommend that the bending moments be based on 
the net span, i.e., the distance between the faces of the columns. 
The following coefficients are recommended: 

Let /; = net span, i.e., distance between the faces of column, ft.; 

w = dead and live load on wall beam, lb. per lin. ft. 

For interior spans the bending moments should be as follows: 

Negative bending moment, M = — ,uwli? ft.-lb. or wl,? in.-lb.; 

Positive bending moment, M = z;wl,? ft.-lb. or 0.75wl;? in.-lb. 

For end spans where the width of the corner column is less than 
one-tenth of the span, 

Negative bending moment, interior column, 


M = — xul,? ft.-lb. or — 1.2wl1? in.-lb. 


380 DESIGN OF FLAT SLAB STRUCTURES 


Negative bending moment, corner columns, 

M = — 3,wl:2 ft-lb. or — 0.75wl,? in.-lb. ; 
Positive bending moment, 

M = },wli? ft.-lb. or wl)? in.-lb. 


For end spans where the end column is more than 14 times the 
depth of the spandrel and more than one-tenth of the span, the same 
bending moments may be used as for interior spans of the wall beam. 
At the end column, then, the same area of negative reinforcement 
must be used as at interior columns. 

For deep beams, the bending moment must be considered as 
resisted by the reinforcement in the beam only, because the tensile 
stresses in the reinforcement of the slab strip parallel to the beam are 
either non-existent or negligible. 

With shallow wall beams, the slab reinforcement parallel to the 
beam resists part of the bending moment in the wall beam. Usually, 
the effect of the slab reinforcement is small and is not worth con- 
sidering. If desirable to consider the slab reinforcement, it should be 
done as in the following example: 


Example 1.—If effective depth of beam isd = 15 in. effective depth of slab, 
d; = 10in.; k = 0.375. Then the depth of neutral axis for the beam is kd = 5.6 
in. The depth of slab below neutral axis for the beam is 15 —5.6 = 9.4 in, 
while for the slab it is 10 — 5.6 = 4.4 in. If the stress in beam reinforcement is 
fs, the stress in slab reinforcement will be in proportion to its distance from the 


4.4 
neutral axis, or yn f; = 0.47f,. If As is the steel in the strip parallel to the 


beam, the tensile stresses resisted by it are0.47A sfs. To get the bending moment, 
this stress must be multiplied by the distance between the slab steel and the 
center of compression, which is 4.4 + 3kd =4.4 + 3 X 5.6 = 8.1 tinee lhe 
bending moment is M, = 0.47f, X 8.1 = 3. SAebe 

The same area of steel, if placed in the beam, would resist a bending moment 
M. =AcfsX34X15 =13.2Asf.. Comparing this with the above value, it is 
evident that the effectiveness of the slab steel is equal only to 0.27 of the effect- 
iveness of the same reinforcement placed in the beam. 


Requirements of Building Codes.—The requirement adopted by 
the authors, as to the amount of slab load carried by the beam, is 
incorporated in most modern codes. Some codes, however, still 
permit the wall beam to be designed for the wall load only. ‘This 
always gives unsafe results. The Code of the City of Boston requires 
that the wall beam be designed to carry the wall load and, in addi- 
tion, a slab load on a width of slab equal to 20 per cent of the clear 


WALL BEAMS 381 


span of the spandrel beam (irrespective of the magnitude of the span 
of the slab at right angles to the beam). This requirement is safe 
in some instances and unsafe in others. In cases where the span of 
the slab at right angles to the wall is small, the requirement gives 
absurd results. 

1924 Joint Committee Requirements—The 1924 Joint Committee 
distinguishes between a beam having (a) a depth greater than the 
depth of drop panel into which it frames, and one having (b) a depth 
equal to or smaller than the depth of drop panel. The beam (a) 
(deeper than drop panel) is required to be designed for all directly 
superimposed loads plus one-fourth of the distributed load for which 
the adjacent panel (or panels) is designed. Each column strip 

















‘prop ‘Slab Steel” ‘ 
Beam Steel ~ 





Pcie “Moment Arm 
i Sy for Slab Steel =8.1" 


Fig. 134.—Stresses in Beam and Slab Reinforcement at Wall Beam. (See p. 380.) 


adjacent to and parallel with the beam is required to be reinforced 
with one-half of the reinforcement which it would have required 
without the beam. The beam (b) (depth equal to or smaller than 
drop panel) is required to be designed for the load directly superim- 
posed upon it, while the flat slab strip parallel with it is to be designed 
to carry all the slab load. 

The requirement for beams (a) (deeper than drop panel) is logical 
and agrees substantially with the authors’ recommendation. The 
requirement for beams (b) is not correct, as clearly explained above 
and on p. 388. This requirement is based on the assumption of full 
working stresses in slab steel. From Fig. 134, it is evident that if 
the slab steel were stressed to full working stresses, the stress in beam 
steel would have to pass the elastic limit. 

Wall Beam below Slab with Concrete Curtain Wall above.—In 
many cases, the wall beam and curtain wall are built of concrete, 


382 DESIGN OF FLAT SLAB STRUCTURES 


but, for reasons connected with construction, it may not be desired 
to build them at the same time or to make provision for the coopera- 
tion of the concrete wall beam below the slab with the concrete cur- 
tain wall above. In such cases, a definite joint is made between the 
curtain wall and the rest of the construction, at the top of the slab 
and at the columns. The concrete curtain wall is then built after the 
frame is completed. To insure weather tightness and to tie the cur- 
tain wall to the beam, dowels are provided in the beam and are then 
imbedded in the curtain wall. Also, recesses coated with asphalt are 
provided, in the column, for the walls, The curtain wall is rein- 
forced for temperature. The amount of this reinforcement will — 
depend upon the extent to which it is desired to avoid cracking. 

' At least 0.25 of one per cent of longitudinal steel is recommended, 

particularly if the concrete in the wall is exposed. Window sills 
should be reinforced with two }-in. round bars. Satisfactory 
results have been obtained in some cases, in 8-in. curtain wall where 
3_in. round bars were spaced 12 in. on centers; but the small saving 
in reinforcement does not compensate for the unsightly appearance 
of the temperature and shrinkage cracks which are liable to occur. 

The concrete curtain wall is not considered as a part of the beam. 
The method of design of the wall beam, in this case, differs from the 
previous case only in that the curtain wall does not need to be con- 
sidered as adding to the loading of the beam. 

Wall Beam above the Slab.—It is often desirable to bring the 
top of the window as close to the bottom of the slab as possible. In 
such cases, the wall beam is placed above the slab, as illustrated in 
Figs. 135 to 137. 

When no drop panel is used, the bottom of the wall is flush with 
the bottom of the slab. When a drop panel is used, the bottom of 
the beam is at the same level as the bottom of the drop panel. 

The design of the wall beam will depend upon the method of con- 
struction. 

Beam and Slab Built at the Same Time—The wall beam shown 
in Fig. 135 should always be built with the slab. The method of 
design will be the same for wall beams below the slab. 

In the design of wall beams shown in Fig. 136, the slab and the 
wall beam, for its full depth (including the column), are built at the 
same time. The beam is designed as fully continuous. The tem- 
perature steel in the beam is extended into the columns, to prevent 
cracks at the junction of the beam and the column. When beams 


WALL BEAMS 383 
are deep, no bent bars are used, and the negative reinforcement at 
the column consists of short bars placed near the top of the beam. 
In such cases, the beam is tied to the columns so that it cannot con- 
tract freely. To prevent cracks due to shrinkage and temperature, 
at least 0.3 per cent of longitudinal steel is needed. 

This method, when the wall beam is properly designed and built, 
gives very satisfactory results. However, it is difficult to construct. 
The formwork for the wall beam is expensive, and difficult to keep 
in place during construction. It is difficult to prevent the concrete 
in the wall above the slab from running out into the slab, especially 


~--- Sil 






~---Brick 





2" Diam. Bolt 


Continuous Angle____- 
Sor Brick Veneer 


Fic. 135.—Wall Beam above Slab. (See p. 382.) 


when the spandrel is so high that considerable pressure is developed. 
Rather dry mix should be used in such cases, and the concrete in the 
wall should be poured after the concrete in the slab has begun to 
stiffen very slightly. 

The same loads and moments may be assumed as in the case of 
beams extending below the slab. The bars in the band parallel to 
the wall may be assumed as effective in resisting the positive bend- 
ing moment in the beam, to the extent of 50 per cent of the required 
steel area in the beam. 

Stirrups should be used throughout the length of the beam. 
They act partly to suspend the slab load from the beam. 

Beam and Slab Built Separately.—In the design shown in Fig. 137 
the portion of the beam above the slab is built after the frame is 


384 DESIGN OF FLAT SLAB STRUCTURES 


completed. To get cooperation of the upper portion with the 
lower portion of the beam, sufficient stirrups are provided throughout 
the beam. Since no provision is made for resisting negative bending 
moment, the wall beam should be designed as simple supported, 
using the distance between faces of the column as the span. The 
design load should be taken as in previous cases. The reinforcement 
should be computed for three-quarters of the total effective depth 
of upper and lower portions of the beam. The fraction is used to 
provide for the possibility of imperfect cooperation between the 
two parts of the beam. With this type of design, the forms are 
often stripped before the upper part.of the beam is built. For 
this reason, the lower portion of the beam must be provided with 


re 


Steel 


Construction 0.21, +12 Negative Moment 
OTIC S0 eam aaa Reinforcement. 











: d. Temperatu 








Note: 
Column Bars are Spliced above. Construction Joint. 
Dowels from column below should extend proper 
length above Construction Joint. 


Construction Joint~. 


Fic. 136.—Concrete Spandrel Built Monolithic with Slab and Columns. (See 
p. 382.) 


sufficient reinforcement to carry safely the dead load of the slab 
without the assistance from the upper part. It is particularly 
important to provide negative bending moment reinforcement in 
the lower part of the beam, as shown in Fig. 137. 

The upper part of the beam should be provided with proper 
amount of temperature reinforcement. Since the beam is able to 
contract to some extent, the amount of temperature steel may be 
smaller than in previous case. The actual amount will depend upon 
conditions. ‘The authors recommend one-quarter of one per cent for 
temperature reinforcement. 

Recesses should be provided in the columns for the upper part 
of the beam. The joint should be water-proofed by using tar felt or 
similar material. 

Sometimes an attempt is made to tie the separately poured upper 
part of beam to the column by providing in the columns horizontal 


WALL BEAMS 385 


dowels and longer horizontal bars near the top of the beam, and 
thus provide continuity for the upper part of the beam when built. 
This is not desirable for construction reasons. It is necessary to 
drill holes in the forms for reinforcement. These must be located 
with care so that the hole in one face is just opposite the hole in the 
other face. After the holes are drilled there is always some difficulty 
in placing the dowels in the forms. F inally, great difficulty is 
experienced in stripping the forms. The sides must be moved 
horizontally to clear the dowels. This movement cannot be con- 
veniently made on account of stirrups extending up from the slab. 
For these reasons, this method is not popular on the job and is not 
advocated. : 


Concrete Sill 


Cast Separately -~, §- : "Rd. 


Recess in Column., 
Horizontal 















Construction Joint Roughen Surface and coat with+ 
neat cement grout before pouring Spandrel. 


Ita. 187.—Concrete Spandrel Built Separately. (See p. 383.) 


Instead of providing horizontal dowels in the columns for the 
wall beams, the Aberthaw Construction Co. of Boston places hori- 
zontal pipe sleeves in the column forms at the proper height. After 
the column forms are stripped, steel bars of proper length, extending 
on both sides of the column, are inserted in the pipe. When imbedded 
in the wall beam, they serve as negative bending moment rein- 
forcement. 


GENERAL REMARKS 


It is impossible to design flat slab construction properly without a 
thorough understanding of the action of continuous structures and 
particularly of continuous flat slabs under varying conditions of 
loading. Rules and formulas given in preceding paragraphs are 
mostly for typical panels. In actual construction, odd panels will 
be found, the designing of which must be left to the judgment. of 
the designer. Sometimes, whole floors consist of an assortment of 


oe DESIGN OF FLAT SLAB STRUCTURES 


odd panels. In all such cases, the designer should be able to judge 
how the various parts of the floor will deflect when loaded. Follow- 
ing general formulas blindly may give unsafe results. 

Another perplexing point is the effect of beams on the action of 
the adjoining slab. This may be such as to require complete change 
in the amount and location of the reinforcement. 

Of importance it is to study the effect of openings on, flat slab 
action. 

Tensile Reinforcement Does Not Transfer Loads to Supports 
but Resists Stresses Due to Bending.—The designer should under- 
stand the function of reinforcement which is often misunderstood. 
There is a notion prevalent that bands of steel bars transfer the 
loads from the slab to the support. This is not the case. The 
function of the tensile reinforcement is to resist tension due to the 
bending moments, while the function of transferring the load from 
one section to another and then to the column is performed by 
concrete through its resistance to shear. While this is obvious 
and should not require comment, in order to correct the prevalent 
wrong notion, full explanation is here given. 

In the center, where the external shear is zero and therefore 1s no 
load to be carried, the tensile stresses in a band of bars is a maximum. 
These stresses decrease gradually until at the point of inflection they 
become zero. At the point of inflection, the effect of the reinforce- 
ment is practically negligible. While the stresses in the tensile 
reinforcement decrease from a maximum to zero, the load to be 
transferred to the column equal to external shear steadily increases 
from zero at the center until at the point of inflection, the load to 
be carried to support amounts to about three-quarters of the total 
load on the panel.. Thus, the effect of the steel is largest where 
there is no load to be transferred to the column and is nil at the 
point of inflection where three-quarters of the load on the panel need 
to be taken care of. This fact proves that the bands of bars do 
not perform the function of transferring the load towards the column. 
If they did, their stresses would increase, instead of decreasing with 
the increase of external shear. T he function of transferring the 
load from section to section is performed by shearing stresses in the 
concrete. In the center, where the load to be carried is zero, the 
shear is zero. With the increase in load to be transferred, the total 
shear increases in proportion until it reaches its maximum at the 
column. 


GENERAL REMARKS O87 


Suspension Action Impossible in Flat Slab Construction.— 
Another misconception, fairly widespread specially among advocates 
of the four-way arrangement of steel, is that bands of bars act as 
sort of suspension bars. Nothing could be farther from the truth. 

Suspension action is always distinguished by the load being carried 
to the support entirely by tension stresses in the suspension bars. 
The stresses in the bars depend upon their sag. The horizontal 
component of the stresses equals the static bending moment divided 
by the area of bars and the amount of the sag. The suspension 
bars are under tensile stresses for their full length, the magnitude of 
which is such that their horizontal component is constant for the 
whole length of the suspension bar. A good example of suspension 
action is furnished by the suspension cables of a suspension bridge. 
The cables are anchored at each end and sag between the piers. 
They form the main means of carrying the load. No compression 
members are existing. The cables are free to elongate for their full 
length. 

No such action takes place in flat slab construction. The stresses. 
in each band of bars are a maximum in the center, then they decrease 
until they are zero at the point of inflection. Near the column, the 
stresses increase from zero at point of inflection to a maximum at 
the support. There is no continuous transference of tensile stresses 
as in suspension cable. The bands of bars may be continuous, but 
their action is not continuous. As a matter of fact, as far as tensile | 
stresses are concerned, the bars in flat slab could be cut at the point 
of inflection without affecting the construction, because the tensile 
stress there is zero. If a cable of a suspension bridge was cut, the 
bridge would collapse. A further difference is that for slabs of 
uniform thickness tke sum of tensile stresses at the support is much 
larger than the sum of tensile stresses in the center. In suspension 
structures, on the other hand, the horizontal component of the tensile 
stresses is constant at all sections. 

If bands of bars acted as suspension bars, the strength of the slab 
_ would be only a fraction of the strength flat slabs actually have. 
Flat slabs action is fully explained on page 328. 

Stresses Cannot Act without Corresponding Deformation.—It 
is of particular importance to remember also that stresses cannot 
exist without corresponding deformation of the materials and deflec- 
tion of the member. Neither can deformations of materials (other 
than shortening or lengthening of members free to move, due to 


388 DESIGN OF FLAT SLAB STRUCTURES 


temperature) and deflection of members exist without stresses. 
Reinforcement cannot be effective in resisting tension when it is 
placed in such a way that it is not able to elongate. 

This is of importance in many Cases in connection with flat slab 
design, particularly in connection with the design of beams in flat 
slab construction. In Fig. 188, page 388, for instance, the bottom 
of the beam is some distance below the bottom of the slab. Some 
building codes permit the beam to be designed only strong enough 
to carry its own weight and the load directly superimposed upon it. 
The slab is assumed to act as if there was no beam, and the slab 
load is assumed as being carried from column to column by the 
strip of slab along the beam. ‘The tensile stresses in the slab are 
supposed to be resisted by the half-band of bars parallel to the 


ikial 








Neutral 





~~ Stress in Slab reinforcement 
‘near beam 





_-d7 Stress in beam reinforcement’ 





Fic. 138.—Action of Beam and Slab Reinforcement in Flat Slab Construction. 
(See p. 388.) 


beam. It should be clear that no such action can take place. The 
slab and the beam must deflect as a unit. The position of the 
neutral axis for both the beam and the slab is governed by the beam. 
The slab reinforcement will be near the neutral axis so that for 
quite a distance from the beam the amount of tensile stresses resisted 
by slab steel will be negligible. Before the slab bars can come 
into full action, the beam would have to fail. The design has 
sufficient amount of reinforcement, but it is placed in such a way 
that it cannot perform its function. The proper method of design- 
ing is to make the beam strong enough to carry not only the brick- 
work but in addition 20 per cent of the slab load. (See p. 379.) 


PROBLEMS IN DESIGNING FLAT SLAB CONSTRUCTION 389 


PROBLEMS IN DESIGNING FLAT SLAB CONSTRUCTION 


Method of Designing.—After the spacing of columns has been 
decided upon, the design of flat slab floors or other plane structures 
may proceed as follows: 

1. Assume dead load of slab based on assumption of its thickness. 
This weight added to the live load and to the weight of floor finish 
gives the total loading for the flat slab. 

2. Select size of column head and dimensions of drop panel. (See 
p. ve 

3. Determine restraint at wall column. (See p. 313.) 

4. Compute required thickness of slab and drop panel, using 
formulas, pp. 8 to 28. If slab of uniform thickness throughout the 
floor is desired, the thickness should be computed for panels for 
which the expected bending moments are largest. 

5. Compute shearing stresses at the column head and at the drop 
panel. (See p. 347.) 

6. Compute bending moments Ef the various design sections for 
interior and exterior panels. (See pp. 332 to 335.) 

7. Compute required areas of steel at the various design sections. 
(See p. 351.) 

8. Select the flat slab system to be used and the desired arrange- 
ment of steel. (See pp. 360 to 369.) Select bars which will supply 
the required amount of positive and negative reinforcement in each 
strip. Decide upon the manner of bending the bars. Compute the 
areas of bent bars effective as negative moment reinforcement, and 
compare this area of reinforcement with the required area. If the 
bent bars are not sufficient, add short bars. 

9. Check bond stresses at the column head. (See p. 351.) 

10. Show on the plan, by sketches details of reinforcement for 
typical panels. Give a designating mark to each band of bars or 
reinforcing unit, and label the floor plan with these marks. In a 
schedule of reinforcement, give the number, length, and type of bars 
in each band or unit. Indicate clearly the length of the bars com- — 
posing each band or unit, and show bending sketches for bent bars. 
If the typical panel is to be used as a general guide for panels of 
different sizes, express the lengths of bars and the points of bending 
in terms of the span. 

11. Show clearly the concrete dimensions, on the floor plan and 
in the schedule. 


390 DESIGN OF FLAT SLAB STRUCTURES 


12. Design wall beams. (See p. 379.) 

13. Check ali columns supporting the flat slab, for strength and 
rigidity. Supply bending moment reinforcement in the exterior 
columns when required. (See pp. 305 to 319.) 


EXAMPLE OF FLAT SLAB DESIGN 


The use of the formulas and recommendations will be illustrated 
by the following example. 


Example 2.—Design an interior and exterior panel of a floor in a flat slab 
building. The dimensions of the panels are 20 by 22 ft., as shown in Fig. 139, 
p. 391. The design live load is 150 lb. per sq. ft. The floor finish consists of 
13 in. of bonded granolithic finish (i.e., finish applied separately). 

Use following stresses in lb. per sq. in., fe = 800, fs = 16000, u = 100, 
v = 60, and ratio n = 15. 

Solution.—The method of solving the problem, as outlined on p. 389, will be 
followed. The reference numbers used in the example correspond to the numbers 
used there. 

1. Assuming thickness of slab as 8 in., the design load is 





Live load ces): ce 6 6 Pe ee 150 lb. 
Slab Iodd ti eee ee ee eee 100 lb. 
12 inifinish....90 eh eee 18 lb. 
Total . ota tae See aes w = 268 lb. per sq. ft. 


29. The diameter of column head is made equal to 0.225 times the average 


22 + 20 
+ x 0.225 = 4.72 ft. Assume 4 ft. 9 in. for the 





span. Therefore, c = 


diameter of column head. A rectangular drop panel will be used, with length of 
sides equal to 0.35 of the sides of the rectangular panel. The dimensions of the 
drop panel are 7 ft. by 7 ft. 6 in., with the long side parallel to the long side of the 
panel. 

At the wall column, a square bracket will be used, with a distance from center 
of column to edge, c: = 2 ft. 5in. Half drop panel will be used at the wall. 

8 The slab is supported by concrete columns having the required rigidity. 
Concrete spandrels are also used. Therefore, the exterior panel may be classed 
as Case 1 (see p. 338). 

4. It is desired to use a slab of uniform thickness throughout, and drop panels 
of the same depth. Therefore, the required thickness will be computed for exterior 
panels, which in this case require the maximum dimensions. 

Thickness of Slab and Drop Panel. (Use Formula (18), p. 339.) 


4.75 
The dimensions arel = 22 ft.,b = 7 ft., w = 268 lb. per sq. ft. For 5 Sey 
= ().22 and for specified stresses from diagram, p. 911, the constant C; = 0.0173. 


1, = 0.0173 X 22 X V268 X 22 41.5 = 10.50 41.5 = 12 in. 


EXAMPLE OF FLAT SLAB DESIGN ool 


Thickness of Slab in the Center. (Use Formula (25), p. 340.) 
Using dimensions given above, and since for : = (0.22 and for specified stresses 
from table, p. 911, Cs = 0.0198, 
i = 0.0198 x 22V268 +1 =7.141=8.1. 
Use 8 in. slab. 



























































































































































































ak Be eerie 2 99'.0% ------- -—--- --»}e- --= ---- - + --- 22+. 0% ---------- 
SS | Column |. Middle strip___,| Twocolumn sirips..|_ _Midale strip ___,| Two column strips 
38 strip short direction short direction short direction short direction 
at . t 
ES | 
Cx 2 
ee 
= n 
Come Q 
<= H ‘SI i 
£3 Cy 2|.S 
eee = a Sirs 
368 | ua - = E Sa 
S | : E 
Be | = : Sed 
S& 5 Ojon CT : 1D 
2& a coo DAL P Ly ° § I 
oS = Ser oeeet SiS | 
& HOGI Pal ass oa Saat 
Ys oe | 
3 S 
£& Se ae rial 
g ES SS Ea 
os eee ee 
Se. AU ae ee ee tl Srp a 
Be H See —th wes 
Ss TT a fois TH LS 
hae ad Dl Sangl Bo se] Bi | 
he Cine Sr a resell) ae et 
a Lae as nN ai) 
rH rarer eat as 
cs ee = = t | 
seuss A 1H eee 
aSSumaae 5 ae 4 
aes ee re — fe c | 
Seeceiaes in SS | 
ae aoe! oS 
soneSauee = grey 
£21 0 Gem Ee; Se al i 
Tiara Er] VS 
fs eas s 
ieee Ht s\® 
Sat LH the 
samen a sus a : g 
E 3 S5n= & : 8 aa 
















Bent Bars { ! 
; Hooked Bars Band Cy ve Str Bars Band Co y 8 BandA&B l 
qs Sa ee k eASNOPIA OF 1 eraampre eS Gane? 























Nore. in plan all bars are shown only in bands Ay, Ar, 8 and Co - 
In other bands positions of outside bars only is indicated . 


Fic. 139.—Details of Flat Slab Design. (See p. 390.) 


5. The points of critical shear are distant t:; — 13 = 12 — 14 = 10.5 from 
column head. The diameter of critical section 1s 


10.5 
e+ 2(t; — 14) = AER =i bite ero aD. 


The circumference is 78 X 3.14 = 245 in. 
The net area of panel outside the critical section is found as follows: 
(GN TEES a ee nr ree oe 20 <. 22:= 440 8a, ft. 


676? 
Area inside of critical section, Te xX 3.14 = 33.3 


Dem OU cee, 5s hi Ea Ma « = aw hee RG 406.7 sq. ft. 


392 DESIGN OF FLAT SLAB STRUCTURES 


Total shear at critical section equal to net area times unit load, 
V = 406.7 X 268 = 109 000 lb. 


The weight of drop panel is neglected. 
The shearing stresses are, 


109 000 48 5 Ib ¢ 
y = ———— _ = 48. . per sq. in. 
245 X i X 10.5 Sha 
The same result may be obtained by using Formula (39), p. 349. Since 
20 
| = 22 ft, = 20, m = 5 = 0.91, o = 6.5 ft, - = 0.296, from table, p. 913, 


the constant Cy». = 0.85. 
Therefore 


l 22 : 
y= Ciw, = 0.85 X 268 X rer 47.7 lb. per sq. in. 
Since the computed stress is less than the allowable stress of 60 Ib. per sq. in.,: 


the thickness is satisfactory. 
6. Bending Moment at Various Design Sections.— 


Interior Panel, W = 268 X 20 X 22 = 118 000 Ib. 


4.75 : 
Long Direction, 1 = 22'ft., l = 20 ft., ; = = 0.22. 
M =1.08 X 118 000 x 22(1 — 2 X 0.22)? = 2 040 000 in.-lb. 
Negative Cae = — 0.54M = — 1100000 in-lb 
Moment \ M, = — 0.08M = —_ 164000 in.-lb. 
Positive 9, 3;= 0.23M = 472 000 in.-lb. 
Moment |Mi= 0.15M = 306 000 in.-lb. 
Ret d c’ Aa 
Short Direction, | = 20 ft., 11 = 22 ft., 7 Pog = 0.24. 
M = 1.08 X 118000 x 20(1 — 2 X 0.24)? = 1800 000 in.-lb. 
Negative (M, = —0.54M = — 972000 in--lb. | 
Moment \M, = —0.08M = — 144000 in.-lb. 
Positive a = 0.23M = 414 000 in.-lb. 
Moment |M,= 0.15M = 272 000 in.-lb. 


Exterior Panel, W same as for interior panel. 
Long Direction, 


M = same as for interior panel. 


First interior column line, 


Negative ie = — 0.62M = — 1 265 000 in.-ib. 
Moment | M. = —0.12M = — 245000 in.-lb. 


EXAMPLE OF FLAT SLAB DESIGN 393 


Wail column line, 


Negative {Mi = —0.42M = — 857 000 in.-lb. 
Moment | M, = —0.08M = — 163 000 in.-lb. 
Positive en i 0.27M = 551 000 in.-lb. 
Moment | M, = 0.18M = 367 000 in.-lb. 


Short Direction. 
Bending moments same as for interior panel because the slab is continuous 
in this direction. 
7. Required Areas of Reinforcement.—The required areas of reinforcement 





are obtained from formula As = For fs = 16000, 7f; = 14000. The 


jaf s 
values of effective depth, d, are as follows: At the column head, d = 12 — 1.5 = 
10.5 in.; in the center of column strips, d = 8 — 1 = 7in.; and in the center of 


middle strips where two layers of steel are used, d = 8 — 1.25 = 6.75 in. 





Required 

Effective Required Number 
Location Bending Moment Area of and Diam- 

Depth 
Steel eter of 
Round Bars 
Interior Panel: 
Long Direction: 

Column strips M, = — 1100000 10.5 7.5 sq. in. 26—3"’ 

Middle strips M,= — 164000 7.0 12 7s, 1in. 6— 3” 

Column strips M; = 472 000 7.0 4.8 sq. in 16—3” 

Middle strips M,= 306 000 G2(D 88 p260s in 11— 3” 

Short Direction: 

Column strips M, = — 972000 10.5 .7 sq. in. 22—3/" 

Middle strips M,= — _ 144000 i200 .5-Sq. In. 5— 3" 

Column strips M;= 414 000 7.0 4.3 sq. in 14—3” 

Middle strips M,= 272 000 6.70 | 2.9-8q. in 10—3” 

Exterior Panel: 
Long Direction: 

First interior Pa a 1 265 000 10.5 8.7 sq. in. 29—3" 
Column line M, = 245 000 7.0 2.5 sq. in. g—3”’ 
Wall column -. = 857 000 | 10.5 5.9 sq. in. 20—3”" 
Column line M, = 163 000 7.0 1.7 sq. in. 6—3"" 

Column strip (M; = 551 000 aa 5.7 sq. in. 19— 3” 

Middle strip bps = 367 000 6.75 | 3.9sq.in. | 13—” 








————— ae, 


394 DESIGN OF FLAT SLAB STRUCTURES 


Reinforcement in short direction of exterior panel same as in short direction 
of interior panel. 

8. The two-way system, with the arrangement shown in the first method, 
will be selected (p. 364). (The selection of this system is made for the purpose 
of illustration only and is not intended as a recommendation.) 

Column Strips.—In the column strip, two bars out of three will be bent up 
at both ends and carried across the column into the Udjoining span. In the long 
direction of the interior panel, where sixteen 3-in. rd. bars are used for positive 
bending moment, 10 bars will be bent up and 6 will be carried straight. The 
available reinforcement at the column will consist of 10 bent up bars plus 10 bars 
extended from the adjoining span, making 20 bars in all. Since 26 bars are 
required, 6 short bars must be added. In the short direction, 9 bars out of 14 
will be bent up, making 18 bars available at the column. Since 22 bars are 
required, 4 short bars must be added to complete the required reinforcement. 

In the exterior panel, 12 bars will be bent up and 7 bars carried straight. 
The available reinforcement at the column head consists of 12 bent bars plus 
10 bars extended from the interior panel, or 22 bars in all. Since 29 bars are 
required, 7 straight bars must be added. At the wall column, 20 bars are 
required and only 12 available. Therefore, 8 short bars, provided with a hook 
at the wall end, must be added. 

In the short direction, the same arrangement of steel should be used as in the 
interior panel. 

Middle Strips.—In the middle strip, one bar in three will be bent up at both 
ends and extended into the adjoining panel. The amount of negative reinforce- 
ment will then be equal to two-thirds of the amount of the positive reinforcement. 

9. Bond stresses will be computed at the column head of interior panels, 
according to formula given on p. 351. 

V =0.35W = 0.35 X 118 000 = 41 300 lb. 

The perimeter of one §-in. rd. bar is 1.96 in., therefore the perimeter of all 
bars effective at the column is Zo = 1.96 X 22 = 48.2. 

41 200 
A382 KE IONS 

This is somewhat higher than the bond stress allowed for deformed. bars. 
To reduce the bond stress to 100 Ib., the number of bars would have to be raulti- 
plied by 194, or bars of smaller diameter used. 

10. The details of reinforcement and the length of bars are shown in Fig. 139, 
p. 391. 

11. Concrete dimensions are given in Fig. 139, p. 391. 

12. The gross span of the wall beam is 20 ft. With wall columns 3 ft. wide, 
the net span becomes 17 ft. 

In this design, the wall beam will be placed under the slab. 

The load on the beam equals: 


U 


= 104 lb. per sq. in. 


Load from slab. 2, ...0.2 22 & 268 = 170 paper 
Spatid?tel jc.) sane 3 125 = 375 lb. per lin. ft. 
Glass "and Sash ise oe 7 eee aa 20 Ib. per lin. ft. 


Weight of beam below slab (assumed) 150 Ib. per lin. ft. 
1 715 lb. per lin. ft. 


EXAMPLE OF FLAT SLAB DESIGN 395 


The external shear equals 


17.0 
EE te ieee = 14 700 lb. 


The bending moment is computed as recommended on p. 379. 

Negative bending moment, M = 1 725 X 17.0? = 499 000 in.-!b. 

Positive bending moment, M = 0.75 X 1725 X 17.02 = 374 000 in.-lb. 
The smallest cross section for the beam, as determined by allowable shear, is 


14 500 
C0g Se 138.8. In. 
z X 120 i 
The depth of the beam is usually governed by the allowable distance from the 
top of the window to the bottom of the slab. It is usually advisable to make the 
beam as shallow as possible, so as to bring the window close to the under side of the 
slab. In this case, assume depth of beam h = 18 in. andd = 18 — 2 = 16. 
Minimum Width of Stem Governed by Shear.—Since, as computed before, 
minimum bd = 138, with d = 16, b = 438 = 8.7 in. It is desirable to use a 
larger value, b = 12, to get larger resistance to torsion. 
Areas of Steel.—Since d = 16 in., f, = 16000 lb. per sq. in., the areas of 
steel required by the previously computed bending moment are ' 
At the column, 


Pres. 405,0007) 7) 
ae 1 16.000. ak 
In the center, 
372 000 
re 1.7 5q. in. 
errata Cee 


Use four 3-in. rd. bars for positive bending moment reinforcement, the area of 
which is 4 X 0.44 = 1.76 sq. in. Bend two bars and extend them over the 
support. The available reinforcement at the column, then, equals 1.76 sq. in. 
Since the required area there is 2.2 sq. in., add one 3-in. rd. bar. The total area 
then is 1.76 + 0.44 = 2.2 sq. in. The straight bars will be extended into the 
column and will be available as compression reinforcement, with an area equal to 
2x 0.44 = 0.88. 

Compression Stresses at the Support.—The bending moment at the support 
is 495000 in.-lb. The dimensions are 6 = 12 in., d=16in. The area of 


252 
tensile reinforcement is Asi = 2.2 sq. in., pi = Cen = 0.0115. The area 
' ; 0.88 
of compression reinforcement is A’; = 0.88 sq. in., p’ = eas = 0.0046. 


Since, with f, = 900 and f,; = 16 000, for a balanced design the ratio of rein- 
forcement is p = 0.0129, which is larger than the ratio used, the compression 
stresses are smaller than allowable and the design is satisfactory. 

Diagonal Tension Reinforcement.—Diagonal tension reinforcement may be 
computed as in ordinary design. However, the authors advocate the use of a 
larger number of stirrups near the ends of the beam than required by diagonal 
tension, because there they may assist the beam in resisting torsional stresses. 


396 DESIGN OF FLAT SLAB STRUCTURES 


Joint Committee, 1924, Flat Slab Specifications.—The Joint 
Committee, 1924, flat slab specifications are reprinted as appendix in 
Vol. II of this treatise. 


CHIGAGO AND NEW YORK CITY FLAT SLAB REGULATIONS 
(In effect 1925) 
The regulations of the two cities are given below. 


Definitions. 

Chicago Code: ‘“ Flat slabs as understood by this ruling are reinforced concrete 
slabs, supported directly on reinforced columns with or without column heads 
at the top, the whole construction being hingeless and monolithic without any 
visible beams or girders. The construction may be such as to admit the use of 
hollow panels in the ceiling or smooth ceiling with depressed panels in the floor.” 

New York Code: ‘The rules governing the design of reinforced concrete 
flat slabs shall apply to such floors and roofs, consisting of three or more rows of 
slabs, without beams or girders, supported on columns, the construction being 
continuous over the columns and forming with them a monolithic structure. 

“Por structures having a width of less than three rows of slabs or in which 
irregular or special panels are used and for which the rules given below do not 
apply, the computations and the analysis shall, when so required, be filed with 
the superintendent of buildings.”’ 


Notation. ) 
1 = distance center to center of columns of the side of square panel, or 
average distance of the long and short sides of rectangular panel; 
w = dead and live unit load, lb. per sq. ft.; 


W = total dead and live load on panel in |b. In New York the dead load 
should include weight of drop panel; 
W, = =total live load on panel under consideration. 
COLUMNS 
General Limitations for Interior and Exterior Columns. 
Chicago Code: ‘ The least dimension of any concrete column shall be not 


less than one-twelfth the panel length, nor one-twelfth the clear height of the 
column.” . 

New York Code: “ For columns supporting reinforced concrete flat slabs, 
the least dimension of any column shall be not less than one-fifteenth of the 
average span of any slabs supported by the columns; but in no case shall such 
least dimension of any interior column supporting a floor or roof be less than 
sixteen inches when round nor fourteen inches when square; nor shall the least 
dimension of any exterior column be less than fourteen inches.”’ 


Interior Columns. 

Chicago Code: ‘‘ The interior columns must be analyzed for the worst condi- 
tion of unbalanced loading. It is the intention of.this ruling to cover ordinary 
eases of eccentric loads on the columns by the above requirement. Where the | 
minimum size of column therein specified is found insufficient, however, the 
effect of the resulting bending moment shall be properly divided between the 


CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 397 


adjoining slab and the columns above and below according to best principles of 
mechanics and the columns enlarged sufficiently to carry the load safely.” 

New York Code: “ Interior columns shall be designed for the bending moments 
developed by unequally loaded panels, eccentric loading or uneven spacing of 
columns. The bending moment resulting from unequally loaded panels shall 
be considered as 1/40W ,L, and shall be resisted by the columns immediately 
above and below the floor line under consideration in direct proportion to the 
values of their ratios of I/h. 

“Roof columns shall be designed to resist the total moment resulting from 
unequally loaded panels.” 


Wall Columns. 
Chicago Code: ‘‘ Wail columns in skeleton construction shall be designed to 


~ 


. : WL WL 
resist a bending moment of "60. at floors and ay at roof. The amount of steel 


required for this moment shall be independent of that required to carry the 
direct load. It shall be placed as near the surfaces of the column as practicable 
on the tension sides, and the rods shall be continuous in crossing from one side 
to another. The length of rods below the base of the capital and above the 
floor line shall be sufficient to develop their strength through bond, but not - 
less than forty diameters, nor less than one-third the clear height between the 
floor line and the base of the column capital.’ (See Fig. 109, p. 318.) 

New York Code: ‘“ Wall columns shall be designed to resist bending in the 
same manner as interior columns, except that W shall be substituted for Wi 
in the formula for the moment. The moment so computed may be reduced 
by the counter moment of the weight of the structure which projects beyond the 
center line of the wall columns.”’ 

Slab Thickness.—The thickness of slab must not be smaller than 5/1 for 
floors and 751 for roofs. Absolute minimum is 6 in., except that New York 
allows 5 in. for roofs. 

Formulas for determining thickness of slab for the loading: 


Chicago f= VW for all cases 
New York ¢ = 0.024IV'w + 13 for slab without drop panels 
t = 0.021V w + 1 for slabs with drop panels. 


Effective Depth of Slab.—The effective depth of slab is the distance from the 
center of gravity of the effective reinforcement to the compression face of the 
slab. A protective covering for the reinforcement equal to 1 in., measured 
from outside face of the lowest layer of steel, shall be provided. 

Column Heads.—Effective diameter of column: head is as defined and de- 
scribed on page 319. Its diameter must not be less than 0.2251 for both codes. 

Drop Panels.—Drop panels are as defined and described on page 322. Thick- 
ness and width of the drop panel must be sufficient to resist shearing stresses at 
the column head and at the edge of drop panel and compression stresses at the 
column head. In square panels minimum length of drop must not be smaller 
than 3l. In rectangular panels New York Code permits square drop panels 
with length of side equal to } average span; while Chicago requires a rectangular 
drop with length in each direction equal to 4 the span in that direction. 


398 DESIGN OF FLAT SLAB STRUCTURES 


At wall panels drop panels should extend at right angles to the wall from 
center of the column a distance equal to ¢l. 

Shearing Stresses.—The allowable shearing unit stresses for 2 000 lb. con- 
crete shall not exceed following values: At the perimeter of the column head 
120 lb. per sq. in. and at the perimeter of the drop panel 60 lb. per sq. in. The 


V 
shearing stresses should be computed according to formula v = bid except that 
H 


V 
according to New York Code at the column head formula v = bd may be used. 


BENDING MOMENT COEFFICIENTS 


Design Strips and Bending Moment Sections.—The slab is divided into 
design strips as explained on page 330 and the sections of critical bending moment 
correspond to sections shown in Figs. 118 and 119. 

Chicago Code. Square or Nearly Square Interior Panels.—This rule applies 
to panels in which length does not exceed breadth by more than 5 per cent. 


oS ee a SS ES 





| 





Two-Way » Four-Way 
System System 
Column Strips Positive Bending Moment eoWl soWl 
Negative Bending Moment soWl soWl 
Middle Strip Positive Bending Moment zaaWl * aso Wl 
Negative Bending Moment za0Wl za0Wl 








* The reinforcement required by this moment is the area of steel in each of the two diagonal 
bands, and not the effective area at this section. The effective area of both diagonal bands, 
as customarily computed, is 42 per cent larger than the area required by this moment. 

Any other system of reinforcement in which the reinforcing bars are placed 
in circular, concentric rings and radial bars, or systems with steel rods arranged 
in any manner, whatsoever, shall comply with the requirements of either the 
two-way or the four-way system. 


Chicago Code: Wall Panels. 

Wall Panels in Skeleton Construction.—In wall panels supported by wall 
columns and wall beams, in strips at right angles to the wall increase positive 
bending moment by 25 per cent. . 

Wall Panels Supported on New Brick Walls.—Where wall panels are carried 
on new brick walls, thése shall be laid in Portland cement mortar and shall be 
stiffened with pilasters as follows: If a 16-inch wall is used, it shall have a 4-inch 
pilaster. If a 12-inch wall is used, it shall have an 8-inch pilaster. The length 
of pilasters shall be not less than the diameter of the column, nor less than one- 
eighth of the distance between pilasters. The pilasters shall be located opposite 
the columns as nearly as practicable, and shall be corbeled out 4 inches at the 
top, starting at the level of the base of the column capital. Not less than 8-inch 
bearing shall be provided for the slab, the full length of wall. 

The coefficients of bending moments required for these panels shall be the 
same as those for the interior panels except as: The positive bending moments 
at right angles to the wall shall be increased 50 per cent. 


CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 399 


Wall Panels Supported on Old Brick Walls.—Where wall panels are supported 
on old brick walls there shall be columns with standard drops and capitals built 
against the wall which shall be tied to the same in an approved manner, and 
at least an 8-inch bearing provided for the slab, the full length. The coefficients 
should be same as specified for slabs on new brick walls. Where it is impossible 
to utilize the old brick wall in the manner described above, wall columns and 
wall beams should be provided as in skeleton construction and the panel treated 
as provided for skeleton construction. 


Chicago Code: Rectangular Panels. 

This applies to rectangular panels where length of panel exceeds breadth by 
more than 5 per cent but not more than 33.3 per cent. 

In columns strips of two- and four-way system use in each direction same 
bending moment as would be required by a square panel whose span length is 
equal to the length of the respective side of the rectangle. The amount of steel 
in the short direction, however, shall not be less than two-thirds of that required 
in the long direction. 

In the four-way system the amount of steel in the middle strip, positive and 
negative, shall be the same as that required for similar strip in a square panel 
whose length is equal to the mean of the long and the short side of the rectangular, 
panel. 

In the two-way system the amount of steel in the middle strip, positive and 
negative, running in short direction, shall be equal to that required for the same 
strip in a square panel whose length equals the long side of the rectangular panel. 

The amount of steel in the middle strip, long direction, positive and negative, 
. shall be equal to that required for the same strip in a square panel, whose length 
equals the short side of the rectangular panel. 

In no case shall the amount of steel in long direction be less than two-thirds 
of that in the short direction. 


New York Code: Square or Nearly Square Panels. 





Two-Way Four-Way 
System System 





Slab with Drop Panel 


Column Strips Positive Bending Moment gyW! Wl 








100 
Negative Bending Moment gyWl 35Wl 
Middle Strips Positive Bending Moment Wl TW! 
Negative Bending Moment zisW! zeal 








Slab without Drop Panel 





Column Strips Positive Bending Moment gsWl aeWl 
Negative Bending Moment 3gWl sgWl 
Middle Strips Positive Bending Moment zasWl Too! 


133 0 
Negative Bending Moment za5Wl za5WIl 








400 DESIGN OF FLAT SLAB STRUCTURES 


New York Code: Rectangular Panel. 

Where the ratio of long span to short span is less than 1.1 the panel may 
be considered as square panel with panel length equal to average panel length. 

When the ratio of panel length is between 1.1 and 1.33 the bending moment 
coefficients specified for interior square panels shall be applied in the following 
manner: 

In two-way systems the negative moments and the positive moment at right 
angles to the long direction shall be determined ‘as for a square panel of a length 
equal to the greater dimension of the rectangular panel; and the corresponding 
moments on the sections at right angles to the short direction shall be determined 
as for a square panel of a length equal to the lesser dimension of the rectangular 
panel. In both cases the load W shall be taken as the load on the rectangular 
panel under consideration. ‘The amount of reinforcement in the short direction 
shall be not less than two-thirds of that in the long direction. 

In four-way systems, for the rectangular bands, the negative moment on 
the column head sections and the positive moment on the outer sections ahall 
be determined in the same manner as indicated for two-way systems. 

For the diagonal bands, the negative moments on the column head and 
midsections and the positive moment on the inner section shall be determined 
as for a square panel of a length equal to the average span of the rectangle. 
The load W shall be taken as the load on the rectangular panel under consid- 
eration. 


New York Code: Wall Panels. 

The negative moments at the first interior row of columns and the positive 
moments at the center of the exterior panels on moment sections parallel to the 
wall shall be increased 20 per cent over those specified above for interior panels. 
The negative moment-on-moment sections at the wall and parallel thereto shall 
be determined by the conditions of restraint, but the negative moment on the 
midsection shall never be considered less than 50 per cent and the negative mo- 
ment on the column head section never less than 80 per cent of the corresponding 
moments at the first interior row of columns. 


Panels without Drop Panel or Capitals or Both. 

Chicago Code: In square’ panels where no column capital or no depressions 
are used the sum total of positive and negative bending moments shall be equal 
to that computed by the following formula: 


Wl Cc c\3 
M = —|]1.538 —4-4+ 4.18{- 
8 | ie s(() | 


where M = numerical sum of positive and negative bending moments, regardless 
of algebraic signs; 
W = total live and dead load on the whole panel; 
1 = length of side of a square panel, c. to c. of columns; 


c : : E 
es ratio of the radius of the column or column capital to panel length, /. 


This total bending moment shall be divided between the positive and the 
negative moments in the same proportion as in the typical square panels for 


’ 


CHICAGO AND NEW YORK CITY FLAT SLAB REGULATION 401 


two-way or four-way systems specified above for interior and wall panels, re- 
spectively. 

New York Code: The general rules apply to flat slabs with or without drop 
panels. Any difference in treatment of the two types of design is indicated in 
each case. No provision is made for flat slabs without column heads. The 
code states that for structures in which column capitals are omitted, special 
computations should be filed with the Superintendent of Buildings. 

Line of Imflection.—Both codes specify that in the design of reinforced con- 
crete flat slab construction, for the purpose of making calculations of the bending 
moments at sections other than the critical section, the line of inflection shall 
be considered as being located one-quarter the distance, center to center, of 
columns, rectangularly and diagonally, from center of columns for panels without 
drops. The New York Code further specifies that for panels without drops a 
fraction one-quarter should be changed to three-tenths. 


REINFORCEMENT 
Tension Reinforcement.—The required tension reinforcement should be 


M 
computed by formula A, = fj’ The effective depth d is the distance of the 


3J 

center of gravity of all effective bars in the section under consideration from 
-the compression face of slab. The rules regarding the effective reinforcement 
are same as given on page 354, except that in the Chicago Code the area of 
positive reinforcement in the middle strips, as determined from the specified 
bending, is the area of bars in each diagonal band and not the effective area on 
the moment section in the strip. The effective area equals the area in both 
diagonal bands multiplied by the sine of the angle of inclination. This should 
be kept in mind when comparing the total amount of effective steel as required 
by the Chicago Code in four-way and two-way systems as well as in comparing 
with that required by other codes. 


ALLOWABLE STRESSES 


Allowable Stresses in Steel.—The allowable stress in steel in New York 
Code is f, = 16 000 lb. per sq. in. irrespective of the character of steel used. 
In Chicago Code the allowable stress for medium steel is f; = 18 000 lb. per sq. in. 
It should be noted that to allow for this difference in allowable steel stresses, the 
bending moment coefficients in the New York Code were made proportionally 
smaller, so that the areas of reinforcement as computed according to New York 
Code with f, = 16 000 are practically the same as obtained by using Chicago 
coefficients and stresses f; = 18 000. 

Compression Stresses.—Compression stresses in any section of the slab 
may be computed by ordinary beam formulas using the specified bending moment 
for the section under consideration and for width of beam, the width of the 
section. | 

In computing the compression stresses at the column head in slab with drop 
panels the width of beam in the formula should be taken as equal to the width 
of the drop panel. 

Chicago Code provides that if compression area at the column head is insuf- 


402 DESIGN OF FLAT SLAB STRUCTURES 


ficient, it may be increased by introducing compression steel in the bottom of 
slab. 

The allowable unit stresses should not exceed the stresses allowed in 
beam design. At the column head same increase in stresses is allowed as in 
beams at the support. (Namely, 750 lb. per sq. in. for 2000 lb.concrete. ) 


Special Provisions Regarding Flat Slab Reinforcement. 

Splices in bars may be made wherever convenient, but preferably at points 
of minimum stress. The length of splice beyond the center point, in each direc- 
tion, shall not be less than forty diameters of the bars, nor less than two feet. 
The splicing of adjacent bars shall be avoided as far as possible. 

New York Code has this additional provision: ‘“ When the reinforcement is 
arranged in bands at least 50 per cent of the bars in any band shall be of a length 
not less than the distance center to center of columns measured rectangularly 
and diagonally; no bars used as positive reinforcement shall be of a length less 
han half the panel length plus forty bar diameters for cross bands, or less than 
seven-tenths of the panel length plus forty bar diameters for diagonal bands, 
and no bars used as negative reinforcement shall be of a length less than half the 
panel length. All reinforcement framing perpendicular to the wall in exterior 
panels shall extend to the outer edge of the panel and shall be hooked or other- 
wise anchored. 

‘““ Adequate means shall be provided for properly maintaining all slab rein- . 
forcement in the position assumed by the computations.” 

Chicago Code provides as follows: . ‘‘ In order that the slab bars shall be main- 
tained in the position shown in the design during the work of pouring the slab, 
spacers and supports shall be provided satisfactory to the Commissioner of 
Buildings. All bars shall be secured in place at intersections by wire or other 
metal fastenings. In no case shall the spacing of the bars exceed nine inches. 
The steel to resist the negative moment in each middle strip shall extend one- 
quarter of the panel length beyond the center line of the columns in both directions. 

“Slab bars which are lapped over the column, the sectional area of both 
being included in the calculations for negative moment, shall extend not less 
than twenty-five one-hundredths of the panel length for cross-bands, and thirty- 
five one-hundredths of the panel length for diagonal bands, beyond the column 
center.” 


WALLS AND OPENINGS 


Chicago Code provides: ‘‘ Girders and beams shall be constructed under walls, 
around openings and to carry concentrated loads. 

‘The spandrel beams or girders shall, in addition to their own weight and 
the weight of the spandrel wall, be assumed to carry 20 per cent of the wall panel 
load uniformly distributed upon them.” 

New York Code states: “In the design and construction of reinforced con- 
crete flat slabs, additional slab thickness, girders or beams shall be provided 
to carry any walls or concentrated loads in addition to the specified uniform live 
and dead loads. Such girders or beams shall be assumed to carry 20 per cent 
of the total live and dead panel load in addition to the wall load. Beams shall 
also be provided in case openings in the floor reduce the working strength of the 
slab below the prescribed carrying capacity.” 


CHAPTER VII 
CONCRETE AND REINFORCED CONCRETE COLUMNS 


This chapter gives formulas for plain concrete columns (p. 403), 
columns with longitudinal bars (p. 405), columns with spiral rein- 
forcement (p. 419), and columns reinforced with structural steel. 
In each case the authors’ recommendations for design, as well as 
the requirements of various building departments, are given. 

Details of design for columns with longitudinal steel (p. 412), 
for spiral columns (p. 432), and for columns with structural steel 
(p. 440) are presented and thoroughly discussed. | 

Particular attention is called to the section entitled ‘‘ Economies 
in Column Design,” p. 445. In this section, factors affecting the 
economy of reinforced concrete columns are discussed, and recom- 
mendations are given as to the most economical mix of concrete and 
the most economical proportions of steel to concrete. 

Reduction of live load to be used in column design is given on 
p. 452. 

Recommendations for columns subjected to bending are given 
on p. 458. 

The bending moments in columns carrying crane loads are given 
on p. 464. 


PLAIN CONCRETE COLUMNS AND PIERS 


Columns or piers, the unsupported length of which does not exceed 
four times the least lateral dimension, may be built of concrete 
without reinforcement. The load to be carried by the columns 
should preferably be centrally applied. If the load is eccentric, for 
rectangular or square columns the eccentricity must be less than one- 
sixth of the dimension of the column in the direction of the eccen- 
tricity. For larger eccentricity, the column must be reinforced. 

For central loads, the required area equals the total load divided 
by the allowable unit stress. This and other requirements may be 
expressed by the formulas below. 

403 


404. CONCRETE AND REINFORCED CONCRETE COLUMNS 


| 


Let P = total central load in Ib.; 
A = effective area of cross section of column in sq. 1n.; 


f, = compressive unit stress in concrete in lb. per sq. in. 
Then 

Total Load, 

P = Af,: seat 2 
Required Effective Area, 

en 
Stress in Concrete, 

BP 
ey Ge 


In the above formulas the same units of area must be used for A 
as for fe. In the notation, it was assumed that the area, A, is in 
sq. in. and the stress, fe, in lb. per sq. in. Therefore, the load, P, 
is in Ib. If the area, A, is in sq. ft., and the load, P, in lb., the 
stress, fe, will be in lb. per sq. fis: 

Effective area is the area within fireproofing as discussed on 
p. 272. For piers imbedded in earth, such as pedestals above the 
footings, the total area may be considered as effective, for plain 
columns not exceeding in height four times the least lateral dimen- 
sion, as in such cases no fireproofing 1s required. Allowable unit 
stresses are same as given in table on p. 407. 

Formulas for eccentrically loaded columns or piers are given on 
Dory: 


REINFORCED CONCRETE COLUMNS 


Reinforced concrete columns may be divided according to the 
method of reinforcement into: 

Columns with vertical bars only; 

Columns with vertical bars and continuous closely spaced spiral; 

Columns with structural steel core; 

Columns with cast-iron core. 

The columns with structural steel and cast-iron cores may also 
be provided with spiral reinforcement. | 

Unsupported Length of Column.—The unsupported length of 
reinforced concrete columns shall be taken as: 

(a) In flat slab construction, the clear distance between the floor 
and under side of the capital; 


1In European practice, P is in kilograms, f; in kilograms per square centi- 
meter, and, consequently, A in square centimeters. 


COLUMNS WITH LONGITUDINAL BARS 405 


(b) In beam-and-slab construction, the clear distance between 
the floor and the under side of the shallowest beam framing into the 
column at the next higher floor level; | 

(c) In floor construction with beams in one direction only, the 
clear distance between floor slabs; 

(d) In columns supported laterally by struts or beams only, the 
clear distance between consecutive pairs (or groups) of struts or 
beams, provided that to be considered an adequate support, two 
such struts or beams shall meet the column at approximately the 
same level and the angle between the two planes formed by the axis 
of the column and the axis of each strut, respectively, is not less than 
75 degrees nor more than 105 degrees. | 

When haunches are used at the junction of beams or struts with 
columns, the clear distance between supports may be considered 
as reduced by two-thirds of the depth 
of the haunch. 

The above recommendations agree 
with Joint Committee, 1924, Specifica- 
tions. : 


Longitudinal Bar 


COLUMNS WITH LONGITUDINAL BARS 





A typical design of a column with 
vertical steel only is shown in Fig. 
140. The formulas given below have 
- been universally accepted for columns 
with longitudinal reinforcement. The Fic. 140.—Typical Design of 
various specifications and Building Column with Vertical Steel only. 
Codes differ only in the allowable unit erga hy | 
stresses and the allowable maximum and minimum amount of steel 
(see table on p. 409). 


Let P = total column load in lb.; 

A = effective area of cross section of column in sq. in.; 

A, = area of cross section of vertical steel in sq. 1n.; 
p =ratio of area of vertical steel to area of concrete 
=A,+A; 

f. = compressive unit stress in concrete, lb. per sq. in.; 
compressive unit stress in steel, lb. per sq. in.; 
f = average compressive unit stress, lb. per sq. in.; 
n = ratio of modulus of elasticity of steel to that of concrete. 


— 
| 


406 CONCRETE AND REINFORCED CONCRETE COLUMNS 


General Formulas. 


Pee ee Aj,|) — ame Tp) tees (4) 
P= Af .where. f.=fi1+ @-— pee oe (5) 


Formulas to be Used in Designing Columns. 
Area of steel for assumed area of concrete, A, and specified 
stress, fc. 


P — Af. 
A, = ——> 7 
in — Df. a 
Area of concrete and steel for assumed percentage of steel, p, 
and specified stress, fc. 
f=fll+ nm —Il)ph ee es 
iB 
A= LO ee 
ee (9) 
Age At 


Values of f for different percentages of steel are given on p. 916. 
Formulas to be Used in Reviewing Designs. 
Safe load for specified unit stress, f., and known dimensions of 
column. 


P=flA 4 


Unit stress in concrete and steel for given column load and known 
dimensions of column. 


P 
Beker 6 


Wares See 


Authors’ Recommendations for Columns with Longitudinal 
Steel Only. 


Effective Area.—Effective area of column is the area within fire- 
proofing, in columns exposed to danger from fire. ! 

In columns underground, also in reservoirs for water, full area 
may be taken as effective. 

In wall columns faced with 4-in. brick laid in Portland cement — 
mortar, the facing may be considered as fireproofing. 





COLUMNS WITH LONGITUDINAL BARS 407 


Percentage of Steel—Minimum, 1 per cent; maximum, 6 per 
cent. 

Where, for architectural reasons, the concrete column is larger 
than required by stresses, the minimum percentage of steel should 
- be based on the area of concrete required by stresses and not by the 
area actually used. See Example 4, p. 410. 

Allowable Unit Stress, fe. 





~ 


Ultimate Allowable Unit Ratio of 
Mix of Concrete Strength f’. Stress, fe Moduli of 
Lb. per Sq. In. Lb. per Sq. In. | Elasticity, n 











Vi et. 2 000 450 15 
lime 13:3 2 500 570 12 
ie? hk s.2 








| 3 000 680 | 10 





Limiting Length of Column.—Unit stresses in above table to be 
used in columns the unsupported length of which is equal to or smaller 
than forty times the least radius of gyration. Radius of gyration 
to be computed for concrete and steel. For longer columns reduce 
stresses as given on p. 434. 

Hoops or Ties.—One-quarter in. hoops or ties, spaced not more 
than 12 in. nor more than 16 diameters of the longitudinal bar. 

Fireproofing.—1% in. from outside face of steel. 


Joint Committee Specifications, 1924. 


Columns to be designed by Formulas (4) to (11). 

Effective Area.—Total area of column is considered as effective. 

Amount of Vertical Steel‘ The amount of longitudinal rein- 
forcement considered in the calculations shall be not more than 2 
per cent, nor less than 0.5 per cent, of the total area of the column. 
The longitudinal reinforcement shall consist of not less than four (4) 
bars of minimum diameter of } in., placed with a clear distance from 
the face of the column of not less than 2 in.” , 

Allowable Unit Stress f, shall not exceed 0.20 f’.. 

Same limitation as to unsupported length of column as given 
above. | 

Lateral Ties. ‘‘ Lateral ties shall be not less than { in. in diameter, 
spaced not more than 8 in. apart.” | 

Fireproofing. (See p. 272.) 


408 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Requirements of Building Codes.—The table on p. 409 gives the 
requirements of the Building Codes of the cities of Boston, Cleveland, 
Chicago, New York, and Philadelphia, as they were in effect in the 
year 1925. 

Design of Columns with Longitudinal Bars Only.—In designing - 
columns with longitudinal bars only, the method of procedure is as 
follows: 

The load to be carried is computed first. The working stresses 
are then selected. Usually these are fixed by Building Codes; if not, 
the stresses recommended by the authors should be used. In a 
column there are two variables, namely, the size of the cross section 
and the amount of reinforcement. In determining the dimensions, 
therefore, it is necessary either to assume the size of the column and 
then compute the required area of steel from the formulas, or to 
assume the ratio of steel, p, and find the area of column and the 
amount of steel. The problem will resolve itself into one of the 
examples given below. | 

Example 1.—Given column load P = 400000 Ib. Design a square column 
of the smallest possible size, using the stresses allowed by the Boston Code. 


Solution.—Since it is desired to get a column of the smallest possible cross 
section, the richest mix of concrete (1:1: 2) and the largest percentage of steel 
(p = 0.04) will be used. The allowable stress for 1 : 1 : 2 concrete isf,; = 742 lb., 
R= 10. 

For p = 0.04, the average stress, fromf = fell + (n — 1)pl, 1s 


f = 742[1 + (10 — 1) x 0.04] = 1019 Ib. per sq. in. 


f 





= 392.5 sq. in. 


The side of the square equals 4/392.5 = 19.8 in. 
Adding 14 in. on each side for fireproofing, the size of column is 19.8 + 3 = 
92.8in. Actually, a 23-in. square column should be used. 
The required area of steel is A = pA = 0.04 < 392)5"="15-7 “sq: in: 
sixteen 1-in. square bars may be used. 
Result: Column 23 in. square; 
Mix tiki tees 
Sixteen l-in. square bars. 
Example 2.—Given column load, P = 400 000 Ib. For architectural reasons, 
square columns are required. T he outside dimension of column must not exceed 
25in. Find required mix of concrete and amount of steel, using Boston Code. 
Solution.—The richest mix of concrete will be used, as it gives the most eco- 
nomical design. From table on p. 409, for 1 : 1 : 2 concrete the allowable stress, 
according to Boston Code, is fe = 742 lb., n = 10. Fireproofing should be 13 
in. thick on each side. 


409 


COLUMNS WITH LONGITUDINAL BARS 


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410 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The effective side of the square column, obtained after deducting fireproofing, 
is 25 —- 2 X 1} =22in. The effective area is A = 22? = 484 sq. in. 
The load carried by concrete alone is Af, = 484 X 742 = 359 100 lb. 
The balance of the load to be carried by vertical bars is 
P — Af, = 400 000 — 359 100 = 40 900 lb. 
Since the allowable stress in steel is (n — 1)fp = 9 X 742 =6 680 lb., the 
required area of steel is 


P — Af, 40900 
= = — = 6 Fs 12 C . 
(ne Ufee 6 680 oe 





s 


Fight 1-in. round bars, giving an area of 6.28 sq. in., may be used. 
Result: 25-in. square column; 
1:1: 2 mix of concrete; 
Wight 1-in. round bars. 


Example 3.—Column load, P = 400 000 lb. Find dimensions of column and 
area of steel, using 1 : 1 : 2 mix of concrete when the amount of steel for economi- 
cal reasons is limited to 1 per cent, using Boston Code. 

Solution. —The allowable stress for 1 : 1 : 2 mix, according to Boston Code, is 
fo = 742 lb. per sq. in., n = 10. 

For 1 per cent of steel, the average stress is 


f =fedl + (n — 1p] = 742[1 + 9 X 0.01] = 809 lb. per sq. in. 
The required effective area of concrete then is 


4 _P. _ 400.000 
of oe BOS 





= 494 sq. in. 


Effective side of square column, 


d = V494 = 22.2 in. 
Required area of steel, 


A, = pA = 0.01 X 494 = 4.94 sq. in. 


Two 1-in. round bars plus six 7-in. round bars give an area of 5.18 sq. in. 
Since 14 in. fireproofing is required, side of column is 


d+2X1.5 = 22.243 = 25.2 mn. 


Actually the column side should be made 26 in. 
Column 26 in. square; 
175) Bas 
Two 1-in. round bars plus six 3-in. round bars. 


Example 4.—The size of a rectangular wall column, as fixed by architectural 
requirements, is 40 by 16 in. The load is P = 150000 lb. Find the required 
mix of concrete and the amount of vertical steel, using Boston Code. 

Solution.—Allowing 14 in. on each side for fireproofing, the area of the effect- 
ive section is (40 — 3) X (16 — 3) = 481 sq. in. This section is larger than 


COLUMNS WITH LONGITUDINAL BARS 411 


required to carry the load. Using 1 : 2: 4, mix, and one per cent of steel, the 
allowable average stress if f = f,[1 + (mn — 1)p] = 495 K 1.14 = 564 lb. per 
sq. in. The required area of concrete to carry a load of 150 000 lb. is 


P _ 150.000 


i 064 


The area of vertical steel will be based upon this required area of concrete. 
Therefore, 





= 266 sq. in. 


A, = 0.01 X 266 = 2.66 sq. in. 
Result: Use 1 : 2 : 4 mix, and six 3-in round bars. 


Reviewing of Column.—In reviewing columns already designed, 
two problems may present themselves. First, the dimensions of the 
column may be known, and the stresses may have to be computed 
for a given column load. Second, for known dimensions of column, 
it may be required to find the maximum load the column is able to 
carry. 

Often the reviewer finds that the design is unsatisfactory and he is 
called upon the change the dimensions. In such a case he may retain 
the size of column and increase or decrease the amount of steel, he 
may change the mix, or he may change the whole design. In any 
case, the methods given for the design are used. 


Example 5.—Dimensions and design of the panel are given. Live load is 
specified. From these the computed load on column is 300 000 lb. Dimensions 
of column are 24 in. square. Reinforcement consists of eight 1-in. round bars. 
Required fireproofing, 2 in.; mix of concrete, 1 : 13 : 3, n=12. 

Determine the stresses in concrete and steel. 

Solution.—With fireproofing of 2 in., the effective area of column is A= 
(24 — 2 X 2)? = 20? = 400 sq. in. The area of steel is A; = 8 X 0.785 = 6.28 
sq.in. Hence, A + (n — 1)As = 400 + 11 X 6.28 = 469.1 sq. in. 

P 





The stress in concrete ore Formula. (11) oj. ene AL is f, = 
300 000 630 1h . 
= . per sq. in. 
469.1 on gees 


The stresses in steel are f, = 639 X 12 = 7670 lb. per sq. in. 
This example may also be solved by using Table 21 on p. 916. The value 





of p is computed, p= Hee = 0.0157. Then the average stress, f, is found, 
f an = 750 lb. per sq. in. For f = 750 and p = 0.0157, the stress, f;, is 
found from the table by interpolation. 

Result: 


fs = 639 lb. per sq. in. 
fs = 7670 lb. per sq. in. 


412 CONCRETE AND REINFORCED CONCRETE COLUMNS 


DETAILS OF COLUMNS WITH. LONGITUDINAL STEEL ONLY 


General Requirements.—Steel bars used for column reinforce- 
ment must be straight. They must be placed in vertical position 
and be prevented from displacement by hoops or ties. In square and 
rectangular columns, the bars should be arranged symmetrically. 
They should preferably be placed along the perimeter of the effective 
area. Since vertical bars, if properly imbedded and placed sym- 
metrically, take their stress regardless of their location in the column, 
the effective area of the column (contrary to action in a spiral column) 
has no relation to the area of concrete located between the lines of 
steel. Thus, a requirement of 14 in. of fireproofing does not mean 
that the steel bars must be placed within 1} in. of the surface. On 
the contrary, the bars should be placed far enough away from the 
forms to give ample room for placing concrete. A minimum dis- 
tance of 2 in. is recommended. 

Sometimes, when a large number of bars is used and it is not 
possible to accommodate all of them along the perimeter, some bars 
are placed in the middle section of the column. This is not advisable, 
however, as the bars in the central part are hard to keep in place. 
They also interfere with the placing of the concrete. 

To get a symmetrical arrangement of reinforcement, an even 
number of bars should be used in a square or rectangular column. It 
is preferable that all the bars used in a column be of the same diameter. 
In many instances this is not possible without considerable waste of 
steel. Then the column reinforcement is made up of two groups 
of bars (each consisting of an even number of bars for square or 
rectangular columns). The bars in the two groups should vary in 
diameter by not more than one-eighth in. Thus, to get.an area of 
steel equal to 6.9 sq. in., five $-in. round bars plus five 1-in. round 
bars may be used. To use ten 1-in. round bars would mean a waste 
of 3.13 lb. of steel per lineal foot of column, which for a large number 
of columns may amount to a considerable tonnage. Seven 1-in. 
square bars give a close enough area and may be used in round 
columns. They cannot be used in square or rectangular columns, 
however, because it would not be possible to arrange them sym- 
metrically. 

There is one objection to using bars of different diameters in 
one column. Where experienced labor and good supervision are not 
available, there is danger that all small sizes may be used in one 


COLUMNS WITH LONGITUDINAL BARS 413 


column and all large sizes in another. Under such conditions, bars 
of the same diameter should be used even if it means some waste of 
material. 

At least four bars should be used per column. In good practice, 
the bars should be at least 2 in. rd. to give sufficient lateral strength 
to the columns to withstand any bending which may come on them, 
either by unequal loading or by some accidental horizontal pressure. 
In long buildings, the columns in the end row of panels are sub- 
jected to considerable bending due to temperature changes. The 
building lengthens or shortens from the middle toward the ends, so 
that the top of the end columns must deflect by an amount equal 
to the total expansion or contraction of one-half of the length of the 
building. The bending stresses produced thereby may be large, 
particularly in top columns. To prevent cracks, additional steel 
should be used. 

When a large percentage of steel is used, it is advisable to use bars 
of large diameter so that the clear spacing along the perimeter 
between the bars is not smaller than 3 in. When the reinforce- 
ment of the column consists of very heavy bars, it is advisable, for 
practical reasons, to use at least four light bars. Their use is ex- 
plained under ‘‘ Assembling of Bars.” 

Lapping of Bars.—The stresses from the bars in the upper column 
should be transferred to the bars of the lower column by lapping of 
the bars. For this purpose, the bars from the lower column are 
extended into the upper column a sufficient distance to develop in 
them, by bond, the stresses carried by the bars of the upper column. 
(Lapping by extending the bars from the upper column into the lower 
column is not practicable, because the upper bars would have to be 
placed before the floor system below was poured and they could not 
be kept in position without considerable difficulty.) 

The length of the lap depends upon the stresses carried by the 
bars. For columns subjected to vertical loading only, the length 
of lap is determined by compression stresses in the bar. For columns 
subjected to bending, the length of lap may be fixed by tensile 
stresses. In wall columns, for instance, the full tensile strength of 
the bar should be developed by the lap. 

The length of lap may be taken from the table below, for the 
stresses to be developed and the diameter of the bar. The following 
simple rule may also be useful. For a bond stress of 100 lb. per sq. in. 
the required length of lap is equal to 23 diameters of the bar for each 


414 CONCRETE AND REINFORCED CONCRETE COLUMNS 


1 000 lb. of unit stresses in the bar. For a bond stress of 80 lb. per 
sq. in., the required length of lap is equal to 3g diameters of the bar 
for each 1 000 lb. of unit stress in the bar. 


Length of Lap for Different Stresses in Stee: and Diameters of Bar 
Length Apply Either to Tension or Compression 


Based on allowable bond unit stress of 80 Ib. for plain and 100 lb. for 
deformed bars 





Length of Lap in Inches 














Tne Siete Plain Bars Deformed Bars 
Steel, fs 
ng Diameter of Bars * Diameter of Bars * 
2.) 8 | 8) 2) Dal 1g ee eee ee ee 
6 000 O12 14 16 19S cr 8 Go) 11s et eee 
7 000 11.| 141.16 |\19 | 22 1255 “Ol 1h eis tne 
8 000 13 | 16 | 19 | 22 | 25 | 28 | 10 | 13 | 15 | 18 | 20 | 23° 
9 000 14 | 18°] 21°] 25 | 28 | 82 | UP eee ee 20sec 
10 000 16.| 20 |-283 |-27 | 8311-85 ) 134) 16" 19s 2a ease 
11 000 17 | 21 | 26 | 30-| 34 | 39 | 14917 4 2is ees eeaer 
12 000 19 | 23: | 28 | 338 .| 88 | 42° | 1b (19s 2a 5 S2ON eat 
13 000 20 | 25 | 30 | 36 | 41 | 46 | 16 | 20 | 24 | 28 | 33 | 37 
14 000 99 | 97 | 33:1 38 | 44 1.490 18 2226 Rae sanre 
15 000 92 | 99 | 35 | 41 | 47 | 53 | 19 | 23) 285s a esses 
16 000 25. | 31 138 | 44 | 50 1°56) 20) 25) BON eaon eres 
18.000 28 | 35 | 42 | 49 | 56 | 63 | 23 | 28 | 34 | 39 | 45 | Sl 
20 000 21 | 39 | 47 | 55 | 63 | 70 | 25 | 31 | 38 | 44 | 50 |.56 





* Table may be used for round and square bars. 


If the number of bars used in the upper column is smaller than in 
the column below, some saving in steel may be made by extending 
upward only as many bars as there are in the upper column. The 
remaining bars may be stopped just below the top of the slab. Thus, 
if the fifth floor columns are reinforced with twelve {-in. round bars 
and the sixth floor column with eight 3-in. round bars, only eight 
Zin round bars need to be extended into the sixth floor, and the 
reinforcement of the fifth floor column may consist of eight long 


COLUMNS WITH LONGITUDINAL BARS 415 


bars and four short bars. This method should be used only on jobs 
with good supervision and experienced labor, as otherwise confusion 
may arise and some of the columns may get a larger proportion of 
the short bars than others. 

When the number of bars in the upper'column is larger than in 
the lower column, the lower bars are not sufficient to develop all 
the upper bars. ‘Then, in addition to extending all bars into the 
upper column, it is necessary to use extra dowels equal in number 
to the difference between the number of upper bars and lower bars. 
These should be imbedded in the length required by bond in the 
lower columns and should extend the same length into the upper 
column. 

If the diameter of the lower column is larger than that of the 
column above, the bars, if continued straight, would come outside 
of the upper column. To be brought within the effective area of 
the upper column, the bars should be provided with a reverse bend, 
as shown in Fig. 141. This process is sometimes called “ goose- 
necking.” Care should be taken that the inclined portion of the 
bar is placed within the portion of the column restrained by the floor 
system, because if the bend were placed within the unrestrained part 
of the column there would be a tendency for the inclined bar to 
develop an unbalanced horizontal component which is harmful to 
the concrete. While this is elementary and obvious, it is not always 
appreciated by designers. 

In bars of small diameter, a small offset to bring the lower bars in 
line with the upper ones may be provided by “hickying.” Bars 
below # in. may be “‘ hickied ” after the bar is in place, but before 
concrete is poured in lower columns. Bars over ? in. in diameter 
cannot be readily bent when in place and should be bent before vac 
are placed in the form. 

An incorrect method of lapping the bars, often seen on con- 
struction, is shown in Fig. 142. The value of the bars as laps is 
lost, as the pressure from the column cannot readily be transferred 
to the inclined bars. Also, the bars may interfere with the flow of 
concrete during construction and pockets may form, thereby further 
weakening the columns. 

Butting Bars.—If the column is reinforced with longitudinal bars 
of larger diameter, say over 1 in., the stresses from the bars in the 
column above may be transferred to the bars in the column below 
by butting the bars. The ends of the bars must be milled to even 


416 CONCRETE AND REINFORCED CONCRETE COLUMNS 


bearing. When in place, their ends are kept in position by tight- 
fitting pipe sleeves. From the construction standpoint, it is best 
to extend the bars from the column below about 4 in. above the rough 
slab. After the slab is poured, a pipe sleeve 8 in. long is placed 
over the bar and rested on the concrete. The bars for the column 
above are then placed in the pipe sleeve. 

If the number of bars in the column below is larger than in the 
column above, the extra bars may be stopped just below the top of 
the slab. If the number of bars to be butted in the upper column is 
larger than the number of available bars in the column below, dowels 





Fic. 141.—Correct Method of Splicing Fria. 142.—Incorrect Method of Lapping 
Column Bars. (See p. 415.) Column Bars. (See p. 415.) 


of proper length should be placed in the column below, against which 
the bars of the top column are butted. 

When the diameter of the upper column is smaller than that of 
the column below, the lower bars should be bent or goose-necked in 
the manner described in connection with lapped bars, so that the end 
of the lower bars will come directly under the bar above. Some 
designers slant the bars in the lower column for their full length to 
bring the ends in the proper position. T his is obviously wrong. © 

The butt joint is not capable of resisting tension. When the 
columns are subjected to bending, sufficient steel should extend from 
the lower column to the upper column to resist any possible tension. 
Either extra bending steel should be used, or else part of the column 


COLUMNS WITH LONGITUDINAL BARS 417 


steel, sufficient to resist bending stresses, should consist of bars of 
small diameter and these developed by lapping. In cases where the 
bottom column has a larger number of bars than the top column, the 
extra bars not required for butting may be extended into column 
above to provide the required tensile steel. 

Dowels.—The stresses from the column bars are transferred to 
the foundation by means of short bars called dowels. Their total 
length equals double the length required for developing the stresses 
in column bars. One half of the dowel is imbedded in the founda- 
tion and the other half laps the column bars. The number and size 
of the dowels must ‘be same as that of the bars used for the columns 
above. 

If the depth of the foundation is smaller than the length of imbed- 
ment required for the dowels, their ends should be provided with a 
right angle hook, which should be placed above the footing rein- 
forcement. 

Foundation Plates.—Instead of using dowels, the stresses in the 
bars may be transmitted to the concrete by resting the bars on steel 
plates of sufficient dimensions to keep the bearing stresses on con- 
crete within working limits. In such a case, if the column needs to 
be anchored to the foundation, 
special anchor bars should be Bad veda 
provided. sai 

Column Ties—The bars 7? 
should be kept in position by 
column ties. They are usually 
made of }-in. round bars and 
spaced not more than 12 in. on 
centers (see requirements, p. 407). 

The bending details of ties for 
square and rectangular columns 

are shown in Fig. 143, p. 417. me. 143.—Bending Details of Ties for 
In determining total length of tie Square and Rectangular Columns. (See 
allowance should be made for the p. 417.) 

steel in corners. The wire should 

be lapped at a corner. This method is more satisfactory than 
lapping the wire between the corners, as the joint is more rigid 
and the steel bars are better kept in position. 

For corner columns, the ties should be arranged as shown in 
Fig. 144, p. 418. 








If Square, c=b 


Square = 404 I7t. 
TE CR = 2(b+e) +772 


418. CONCRETE AND REINFORCED CONCRETE COLUMNS 


The ties for special columns should be designed with care, their 
purpose being kept in mind. Fig. 149 shows the ties for a special 
column. It consists of two parts, a rectangle and a stirrup. 





Elevation Section 


Fic. 144.—Ties for Corner Columns. (See p. 417.) 


Example of Column with Vertical Steel Only.—Figure 146, p. 419, 
iJlustrates a typical column with vertical steel only as used in the 
new buildings for the Massachusetts Institute of Technology.? All 


the details are clearly shown. 





Section 





Elevation 


Fic. 145.—Ties for Special Columns. (See p. 418.) 


Assembling Column Steel.—The column steel is usually assem- 
bled on horses. The skeleton, consisting of four corner bars and ties, 


2Stone & Webster Co., Builders. Sanford E. Thompson, Consulting 


Engineer. 


SPIRAL COLUMNS 


is assembled first. Two corner bars 
are placed on horses, the proper dis- 
tance apart; the ties are then spaced 
and wired; then the other two corner 
bars are wired. If the weight of the 
total column reinforcement is not too 
great, all bars are wired before plac- 
ing; otherwise, the skeleton of the 
column reinforcement is dropped into 
place first and finally the remaining 
bars are placed and wired. 

When very heavy bars are used, 
say 1j-in. square bars weighing 5.3 lb. 
per lineal foot, the skeleton consisting 
of four bars, when assembled, would 
weigh several hundred pounds. It 
would have to be handled by a number 
of men, and there would be danger of 
damaging forms while placing the 
column steel. For this reason, when 
heavy bars are used for column rein- 
forcement, it is a good scheme to use, 
in making up the required area, four 
light bars, say #2-in. or {-in., to be 
used for the skeleton. A_ skeleton 
composed of such bars can be readily 
handled and at the same time is stiff 
enough to keep its shape until the rest 
of the bars are placed. 


SPIRAL COLUMNS 


Description of Spiral Columns.— 
Spiral columns are columns reinforced 
either with continuous closely spaced 
spiral reinforcement or of closely 
spaced separate hoops placed as near 
as possible to the outside face of the 
column. In practice continuous spiral 
is used in preference to separate hoops. 







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Fic. 146.—Typical 
Vertical Steel Only. 





Column with 
(See p. 418.) 


420 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The spiral reinforcement is based on the principle that concrete 
confined within the spiral is capable of withstanding very much | 
larger stresses than concrete free to spread sidewise. The function 
of spirals, therefore, is to confine the concrete. In performing this 
function the spirals are subjected to tensile stresses. 

Although spiral alone increases the strength of the column it is 
never used in practice without longitudinal reinforcement. There- 





























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Fic. 147.—Spiral Column. (See p. 420.) 


fore, for practical purposes a spiral column is a column reinforced 
with longitudinal reinforcement and with closely spaced continuous 
spiral placed near the outside face of the column. Such column is 
illustrated in Fig. 147, p. 420. 

There is much difference of opinion as to the method of computing 
the strength of columns with spiral reinforcement. ‘The reason for 
this divergence is discussed on p. 82, in connection with the test 
of columns. ‘The method recommended by the authors, as well as 
the methods required by the Joint Committee, and by the New 


SPIRAL COLUMNS 421 


York, Chicago, Cleveland, par Des and Boston Building Codes, 
will be given. 


Let P = total column load in \b.; 

A = area of concrete core within spiral (the diameter of 
which equals the diameter of spiral measured to the 
center of wire) in sq. in.; 

A, = area of cross section of vertical steel in sq. in.; 

p = ratio of area of vertical steel to area of concrete core 
=A,+ A; 

pi = ratio of volume of spiral reinforcement to volume of 
concrete core; 

f. = compressive unit stress in concrete, lb. per sq. in.; 

f’. = ultimate compressive strength of concrete at 28 days, 
based on 6 X 12 in. or 8 X 16 in. cylinder tests, 
Ib. per sq. in.; 

f’. = compressive unit stress in steel, lb. per sq. in.; 

f, = allowable stress in wire composing spiral reinforcement 
(as used by New York Code). 


Authors’ Recommendation for Spiral Columns.—Formulas for 
Strength of Column.—For spiral columns designed according to 
rules given below the formula for strength is the same as for columns 
with vertical steel only. The effect of the spiral is expressed by 
allowing larger unit stresses. 


Formula for Total Load, 


peau Adil: A) fels, ww Ye eeueek es 1012) 
or, when A, = pA, 
feo T GRO ee (eee el eee Cae meer eres ES) 


c These formulas should be used in the same manner as explained 
for columns with vertical steel only (see p. 406). Formulas (4) 
to (11a) apply here also. 


Allowable Unit Stresses. 
The allowable stress in concrete on spiral columns is equal to 
fc = 0.35 f’,. The stress in steel equals n times the concrete stress. 


Spirals are not figured as adding directly to the strength of the col- 
umn. 


422 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The following stresses may be used: 


enn 




















Ultimate Strength, Working Stress, 
Concrete Mix Se: it; n 
Lb. per Sq. In. Lb. per Sq. In. 
hod ye 3 000 | 1 050 10 
1 ees es: 2 500 880 12 
L224 


2 000 700 15 





0 8 a See eee 


Requirement as to the Amount of Reinforcement. 


Vertical reinforcement, not less than 1 per cent, nor more than 
6 per cent of the effective area of column. 3 

Spiral reinforcement, not less than 1 per cent of the volume of 
the concrete enclosed within the spiral. 

Pitch of spiral and other details as recommended under proper 
headings. 

Unsupported length must not exceed 40 times least radius of 
gyration. For larger lengths, use reduction of stresses given on 
p. 435. 

Fireproofing —Two inches for round and octagonal columns. 
For square columns with round spiral, minimum 14 in. 


Boston, Cleveland and Philadelphia Codes 


The formulas for spiral columns are the same as for columns 
without spiral. The amount of spiral reinforcement must not be 
less than 1 per cent of the volume of concrete within the spiral. 
For such columns the effect of the spiral is expressed in the allowable 
unit stress on concrete. 


P = Af. + (n — AS rr 
also when A, = pA, 
PL 4ytit @ = el 


Allowable unit stresses and requirements are given in table 
below. : 


423 


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424 . CONCRETE AND REINFORCED CONCRETE COLUMNS 


New York Code 
Formula for Total Load on Spiral Columns, 


P= A4f,ttin— DAS. + 2Afe 


also 


P= Af[1+(a—Dp+ 23]. Serta 


From which, for known area of core, A, and percentage of spiral pi, 
Area of Vertical Steel, 


ie as A(f. ae 2pifs) 





A; = 18 
in — if os 
For known area of core A and area of steel A,, 
Ratio of Spiral, 
Lee fA aE (n a LAs] (19) 


Pi <a DAT, ie 
Allowable Unit Stresses to be Used in above Formulas: 


The following stresses may be used: 





| 
Allowable Stress, : 








| Concrete Mix lee . n 
| Lb. per Sq. In. 

Loe 500 15 

1: Lass 600 12 





Allowable stress for cold drawn wire for spiral 
f; = 20000 lb. per sq. in. max. 
but not more than 35 per cent of the elastic limit of the wire. 


Requirement as to the Amount of Reinforcement. 

Vertical reinforcement not less than 1 per cent nor more than 
A per cent of concrete core. 

Spiral reinforcement not less than 3 per cent nor more than 2 
per cent. The ratio of spiral, p1, is the volume of spirals 
per lineal foot of column divided by the volume of the enclosed 
concrete in a foot of length of column. 

Pitch of spiral not more than one-sixth of diameter of core not 
more than 3 in. 


SPIRAL COLUMNS 425 


Fireproofing 2 in.° 
Ratio of length of column to least side not more than 15, minimum 
diameter 12 in. 


Chicago Code 


Formulas for Total Load on Spiral Columns, 


Pee A titan —i1)A fi Zann Afs,  .. .. . (20) 
also 
eee ee bp be 2b np 2 rae (21) 
Area of core for assumed values of p and p1, 
Je ada PPLE Scent A) 


fll + (n — 1)p + 23npil 

Area of vertical steel for assumed A and p1, 
a yee P — Af[1 + 23npi] 
; (n — 1)fe 

Allowable Unit Stresses —The values of f, equal one-fourth of the 


ultimate strength of concrete f’... The table below gives the value 
of f, and n for different mixes of concrete: . 


(23) 





Ultimate Strength, Allowable Stress, 


Mix of Concrete fics aa n 
Lb. per Sq. In. Lb. per Sq. in. | 

Meck 5 2 900 725 10 

| EE 2 400 600 12 

hese 734 2 000 500 15 





Requirement as to the Amount of Reinforcement. 

Vertical Reinforcement.—Not less than the percentage of spiral 
nor more than 8 per cent of the core. 

Vertical bars must not be spaced along the circumference of the 
spiral farther apart than 9 in., nor more than one-eighth of the cir- 
cumference. 

Spiral Reinforcement.—Minimum 4 per cent, maximum 1} 
per cent. 

Pitch of Spiral—Maximum 75 diameter of core, but not more 
than 3 in. 

Fireproofing, 2 in. 

Ratio of length of column to least side shall not exceed 12, 


426 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Joint Committee Specifications, 1924. 


Formula for Total Load on Spiral Column, 


P = Af, + (n— 1)pAf., . =) (24) 
or , 

P = Afjl + (» — 1p), . >) Se (24a) 
where 
| f, = 300 + (0.10 + 4p)f’. lb. per sq. in,” aia 


Strength of Concrete, f’c.—The expected strength of concrete, f ee 
upon which the unit stress, fo, depends according to the report, 
depends not only upon the mix of concrete, but also upon the ‘‘slump ”’ 
of concrete during construction. (See Appendix, Vol. II.) 


Requirement as to Amount of Reinforcement. 

Vertical Reinforcement.—Not less than 1 per cent nor more than 
6 per cent of the core. Minimum amount of steel to consist of at 
least six 4-in. bars. 

Spiral—Percentage of spiral not less than 1 of the percentage of 
vertical bars. : 

Pitch of spiral—Not more than § of the diameter of core, but 
not more than 3 in. 

Additional Requirements——Spiral must be held in place firmly 
and true to line by at least three spacer bars. Spiral reinforcement 
shall meet the requirements of the ‘‘ Tentative Specifications for 
Cold-drawn Steel Wire for Concrete Reinforcement ”’ of the Amer- 
ican Society for Testing Materials. 

Fireproofing —Two inches for round column or octagonal col- 
umns. For square columns with round spiral minimum 1% in. 


Use of Formulas for Spiral Columns. 


Boston, Cleveland and Philadelphia.—In these formulas for spiral 
columns, the percentage of spiral is fixed and does not enter into 
computations. The variables are the area of concrete core, A, and 
the amount of vertical bars, As = pA, as for columns with vertical 
pars only. The design of the column, therefore, proceeds in the same 
manner as for columns with vertical steel only, the difference being 
in the unit stresses. 

The problems in this connection and their solution are the same 
as given on pp. 408 to 411 for columns with vertical steel only. 

New York .Code.—In the New York City formula for spiral 
columns, there are three variables: the area of concrete core, A, the 


SPIRAL COLUMNS 427 


percentage of spiral, P;, and the amount of vertical bars, A;. The 
problem of designing is best solved by means of tables given on p. 918. 

If no tables are available, the columns are designed by trial. 
The diameter of column is assumed first, then a ratio of spiral is 
adapted (see p. 432 for relation of p; to pitch of spiral, also tables 
on pp. 930 to 933); foe the amount of vertical steel is com- 
puted from formula 

: Cis a Fase | 

The problem may be also solved by assuming a size of column 

and ratio of vertical steel. Then the ratio of spiral is found from 
ney “Til. 

If the design is unsatisfactory or uneconomical, another assumption 
is made. Knowing spiral ratio p1, the pitch of spiral may be found 
by Formula (35), p. 432, for accepted diameter of wire. The pitch’ 
may also be found from tables on pp. 930 to 933. 

If it is desired to get the smallest possible column, in the equation’ 


P= Aj.|1 Sen 1)p -- air |, substitute for p and p, the largest 


_ allowable values (pi = 0.02 and p = 0.04). Then the smallest 
area is 


(26) 





(27) 


12 sie (28) 


f.|1 + (n — 1)0.04 + 0.04 | 

If it is desired to use a special percentage of vertical bars and 
spiral, substitute the selected values of p and 7, in the above formula 
for P, and solve for A in the same manner as in the previous case. 

To find the stresses for given dimensions of column and given 
load, P. Values of A, p, and p; are known. Value of fs 18 also 
known as it depends upon the known strength of wire. Divide the 
load, P, by area, A, which gives the following relation: 


fi. + (n— lp + 2p | a 


in which all values except f. are known. To find the value of f, 
would require a solution of a second degree equation. To avoid 


this, a value for the ratio, Js may be assumed, and the equation 
: } q 


fr 


428. CONCRETE AND REINFORCED CONCRETE COLUMNS 


solved for f.. The ratio of is then checked, using the value of f, 


just found. If it is near enough to the assumed ratio, the solution is 
correct; otherwise, repeat the performance with a revised ratio. 


Chicago Code.—The method of procedure with the Chicago formula 
for spiral column is the same as with the New York formula. The 
smallest column is obtained in the same manner. Also, the method — 
of procedure is the same if it is desired to use special percentages of 
steel. It must be remembered, however, that the percentage of 
vertical steel must not be less than that of spiral. 

To review a column for known values of A, p, p1, and for spiral 
column load, P. The stress f. is found in formula, 


f.= SS ye te 
° All + (n — 1)p + 23npil 


in which all values, except f., are known. 


(29) 


Joint Committee Rule, 1924.—In the Joint Committee formulas 
for spiral columns there are three variables, namely A, p, and fz. 
The value of fe depends not only upon the quality of concrete, but 
also upon the percentage of steel. The ratio of spirals varies with 
the variation in the ratio of vertical steel. 

To design columns, it is necessary to assume percentages of steel, 
for which the stress, fe, is found from Formula (25). Finally, the 
area of column is found from 

P 
A lta ipo. (30) 

To get minimum size of column, assume the maximum value for p. 

The design is more involved if it is required to find reinforcement 
for an assumed size of column. For instance, if the area of column, 
A, is known, then — 


f+ — Del =9 (31) 


From this expression it is impossible to get directly the value of p, 
as the value of f, is not known, being dependent upon p. To solve 
the problem, two simultaneous equations must be solved, namely 


jl hie Del =p RNR ES 


and 
f, = 300 + (0.10 + 4p)f".. Renmirntons se ote? 


SPIRAL COLUMNS 429 


: P 
In these equations, a and f’, are known and f, and p unknown. 


By eliminating the value of f., we get a second power equation. 
This, solved, gives the percentage of steel, p. 
To review a design for known values of P, A, and p. The stress 


f. is found from 
te 


ese Dol 


This value, however, does not give any information as to the safety 
of the column. To get this, it is necessary to substitute the values of 
fc and p in the formula f, = 300 + (0.10 + 4p)f’. and find the value 
of f’.. Then compare this ultimate strength with the ee ulti- 
inate strength of the concrete. 

From the above it is evident that working with Joint Committee 
formulas is complicated unless tables are available. 


(34) 


OCTAGONAL SPIRAL COLUMNS 


Octagonal columns with spiral reinforcement are sometimes used . 
because some architects prefer them to round columns on account of 
appearance and because it is easier to fit partitions to the flat surface 
of the column. In some instances, they are used because, through 
local conditions, they are more economical than round columns. 

As the spiral is made round, the effective area and the strength of 
an octagonal column are equal to those of a round column of a diam- 
eter equal to the short diameter of the octagon. The gross area of an 


octagonal column equals eee where d is the short diameter of 


the octagon, while the gross area of a circle is ee An octagonal 


4 
column requires, therefore, 5.4 per cent more concrete than a round 
column of the same strength. 

When comparing the cost of round and octagonal columns, in 
additon to the extra cost of materials, the question of formwork 
should be considered. In flat slab work, in addition to the column 
form, the form for the column head must be considered. The rela- 
tive cost varies with local conditions. 





SQUARE SPIRAL COLUMNS 


Tests prove that square spirals are not effective as such, since 
they are not capable of restraining the concrete laterally and prevent- 


430 CONCRETE AND REINFORCED CONCRETE COLUMNS 


ing it from spreading under load. If it is necessary to use spiral rein- 
forcement in square columns, the spiral must be made round and 
the column designed in the same way as a round column of the same 
diameter. The concrete at the edges, which is outside of the spiral, 
is wasted. A square spiral column requires 27 per cent more concrete 
than a round spiral column of the same strength. 

Square columns with spiral reinforcement should be used only 
where for any reasons square columns are desirable or where round 
or octagonal column forms are not easily obtainable. 

The protective covering for the spiral in square columns varies 
+n thickness at different points in a square section. It is largest 
at the corners and smallest in the middle of the sides. This smallest 
thickness of the protective cover may be made } in. smaller than 
required for round columns, where the cover is of uniform thickness, 
but not less than 13 in. This reduces somewhat the excess of con- 
crete in square spiral columns. 


OBLONG SPIRAL COLUMNS 


Sometimes it is desirable to make the clear distance between 
the faces of the columns in one direction as small as possible, while 








Rounded Corners Rounded Corners Rounded Corners 





Square Corners Square Corners 
























aos + _1 —T ee ‘ 
pee ee 8) es wees ES ea a oe oe co oe 
ee ee ig ec) ES a sos cn os 
a ee BS et ee ii ems SE = gem o oo oe ao 
pis Og pe rt _# =ae oo oe 8 tH 
ee iy eee ES: ron a oe Boek & oe 
pee ee ty oe ee ses fom rp Ht 
oo oe a oe Bee Pp} =e Pt =. 
=e et tts 
Bee Se 8g ee ae : a5 ww : . = 
gees GS Bs pee ee 7B Be: ea B oo oe =o 
a ees el Sg pee me Gee SS: Tae 8 coos aoe 
pee Sol ype ae oo: aan oon oe oo 

ee oo ge ese ae ee: i sam Se = 
tt — ‘es Cy eae 4——F rt 

a 6 c 


‘Fic. 148.—Spirals in Oblong Columns. (See p. 431.) 


the clear distance in the direction at right angles is of minor impor- 
tance. In such a case, oblong columns may be used. Under some 
conditions, oblong columns may be spirally reinforced. 


DETAILS OF SPIRAL COLUMNS 431 


The reinforcement of such oblong columns usually consists of 
two or more spirals. ‘The cross section may be either rectangular 
or oval, i.e., with rounded corners. 

To be economical, the relation of the small diameter to the 
large diameter of the column must be such as to permit the design 
shown on Fig. 148, a or b.. In the column shown in Fig. 148c, the 
concrete between the two spirals is outside of the effective areas. 
It does not contribute anything to the strength of the column, 
and is, therefore, wasted. 

The spiral oblong column actually consists of two spiral columns 
of small diameter, and should be designed as if the spiral columns 
were separated. It is obvious, however, that no extra concrete is 
required for protective covering between the spirals, as there the 
concrete 1s not exposed. The effective area in Figs. 148a and c 
equals the area of the two small spiral columns; while in Fig. 148b, 
where the spirals intersect, the actual area should be computed. 

The rules for spiral columns apply equally here. 


DETAILS OF SPIRAL COLUMNS 


Data for Designing Spirals.— 
Let pi = ratio of spiral = volume of spiral divided by volume 
of concrete core; 
A; = area of wire, square inches; 
A’, = equivalent area of vertical steel of same volume as 
ame | 
je 
d = diameter of column core, inches; 
s = pitch of spiral, inches; 
W = weight of spiral per foot of column height, feet. 


the spiral = pi 


Pitch of Spiral, s, for Known Ratio of Spiral, p: or Known 
Area A’s. 





4A, 
= —-. Pee | aah hig he gene sane 
eink (35) 
VES 3.4 Ch...) RGR Pe eto De) 
Ratio of Spiral, p’, for Known Pitch s. 
4A, 





Pi ds” (36) 


432 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Equivalent Area of Vertical Steel of Same Volume as the Spiral. 


Al, = 3.145 As. « er 


Weight of Spiral, W, per foot of Height of Column (not including 
spacers). 


W = 10.95 Ay. . ee POE 


Wire for Spiral—Spirals are usually made of cold-drawn wire 
having tensile strength of at least 80000 lb. per sq. in. The size 
of the wire is designated in the mill by gage numbers instead of by 
the diameter of the bar. The designers, however, use the diameter 
of the bar almost exclusively. The relation between the gage and 
the diameter of the bar is given in the table below: 


Standard Wire Used for Spirals 























Diameter of Actual Weight of Wi 
Gage Numbers Equivaient Diameter Area of Wire, ee Foot hh 
of Wire Round Bar, of Wire, sq. in. P lb oe 
in. in. : 
é 1 0.4900 0.189 0.636 
: G 0.4305 0.145 0.490 
: 3 0.3625 0.103 0.347 
0 5 0.3065 0.074 0.250 
3 1 0.2437 0.047 0.158 








The spiral should be made of a continuous wire. If this is impos- 
sible, the wire must be spliced to develop the full strength of the wire 
by bond. It must be remembered that the spiral acts in tension 
and not in compression. 

Pitch of the Spiral.—The vertical distance, center to center, 
between two adjacent windings of the wire of the spiral is called 
the pitch of the spiral. The maximum pitch is always limited to 
a certain fraction of the core, and also, as an additional limitation, 


DETAILS OF SPIRAL COLUMNS 433 


it must not exceed 3 in. The minimum pitch is seldom specified, 
but it is limited by the practical consideration that the clear dis- 
tance between the wires should be large enough to permit free flow 
of concrete from the core to the outside shell; otherwise, pockets 
might form at the wires, resulting in a porous fireproofing. The 
clear distance between the wires should not be less than 1 in., which 
fixes the minimum pitch for 3-in. spiral at 14 in., and for 3-in. spiral 
at 12 in. 

Spacers.—It is important, in order that the pitch of the wire 
be made uniform, that the spiral should be maintained in a 
vertical position and that it should be secure from displacement 
during pouring of the concrete. For this purpose, the spiral must be 
securely wired to the vertical bars. The best results are obtained 
by using mechanical spacers. 

The spacers consist of small angles, channels, or T-bars, one leg 
of which is notched at proper intervals to receive the spiral wire. 
At least two spacers should be used per column. Larger spirals 
may require three or four spacers. Usually, it is the best practice 
to have the spirals built up in the shop and deliver them to the job 
collapsed. In some cities (notably in New York), this has been 
against the regulation of the labor union. There the wire may come 
to the job coiled, but must be spaced and wired on the job. This 
increases the cost appreciably. 

In estimating the weight of spiral, the weight of the spacer must 
be added to the weight of the wire. The weight of spacers per foot 
depends upon the cross section of the spacer and also upon the 
diameter of the wire, heavier wire requiring heavier spacers. The 
weight per lineal foot of two T-section spacers for 7-IN. wire varies 
from 1.60 to 2.0 lb.; for 3;, and 2-in. wires from 2.0 to 3.0 lb.; for 
heavier wire 3.60 lb. 

Length of Spiral—By length of spiral is meant the vertical 
distance from the top to the bottom of spiral when erected. A good 
rule is to make the length of the spiral 6 in. shorter than the dis- 
tance from top of slab to top of slab. An extra turn of the spiral 
should be added at the top and bottom, for lap. With such design, 
the top of the spiral is well imbedded in the concrete, and at the 
same time there is no danger of any wires extending above the slab 
owing to inaccuracy in manufacture. 

It is true, as it is often argued, that the column is enlarged at the 
floor level by the floor system, especially in flat slab construction 


424. CONCRETE AND REINFORCED CONCRETE COLUMNS 


where the column is enlarged at the top to form the column head. 
This may serve as an argument for stopping the spiral at some point 
within the column head. 

On the other hand, the spiral columns are almost always of rich 
mix, while in the floor and the column head 1: 2: 4 mix is used. 
The saving in the cost of spiral would be small in comparison with 
the advantage of having a continuous spiral column; therefore, the 
rule given above is recommended. 

Cost of Spiral.—The cost of spiral per ton is larger than the cost 
of bars per ton, to allow for the shopwork of bending and assembling. 
Also, the unit cost of drawn wire is larger than the unit cost of bars. 
The difference usually amounts to from $15 to $20 per ton. 

In figuring relative cost of spiral and vertical bars, it should be 
remembered that the length of vertical steel is greater than the 
length of spiral. The length of vertical bars equals the story height 
plus the lap of the bar while the length of spiral equals the story 
height minus 6 in. The cost of the spacers, if used, should be added 
to the cost of the spiral. 

Details of Vertical Bars.—Vertical bars should be extended 
above the top of the rough slab into the upper column, a sufficient 
amount to transmit by bond the compressive stresses from the 
column above to the column below. The remarks on p. 412 in con- 
- nection with vertical bars for columns with longitudinal steel, apply 
here also. 


LONG COLUMNS 


If the ratio of unsupported length of a column to the least side 
of smallest radius of gyration exceeds the values on which the formulas 
given in previous pages are based, the column becomes a long column 
and must be designed with reduced unit stresses. 

The formula given by the 1924 Joint Committee is recommended. 


Let P; = total safe load on long column; 
P = total safe load on short column (F not more than 40) ; 


1 = unsupported length of column in inches; 
R = least radius of gyration of effective area in inches; 
f.1 = allowable unit stress on long column; 
f. = allowable unit stress on short column. 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 485 


Then 
’ | 
R 
marie 2), sli doy etl hte (39) 
and 
R 
ie =f, Wea aa 120 suns ° abe . (40) 


It should be noticed that for : = 40, the expression in the brackets 


becomes unity. 


COLUMNS FOR FLAT SLAB CONSTRUCTION 


Special requirements for columns in flat slab construction are 
given in the chapter on “‘ Design of Flat Slab Structures.”’ 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 


When small structural shapes are used and their area does not 
exceed 6 per cent of the effective cross section of the concrete, the 
column should be treated as a reinforced concrete column and 
designed according to the formulas given for columns with longitudinal 
bars, on p. 406. The angles should be fastened together with lattice 
work or stays, such as are used in structural steel design. The steel 
shapes should be covered with concrete of a minimum thickness of 
2in. To prevent the separation of this fireproofing from the rest 
of the concrete, it is advisable to place outside of the angles 3 in. round 
ties spaced about 18 in. on centers. 

When the structural shapes are designed to resist the major part 
of the column load, the column becomes, in the strict sense, a struc- 
tural steel column imbedded in concrete. Its design differs mate- 
rially from that of the reinforced concrete column. The formulas 
for structural steel columns are not yet fully standardized and differ 
in building rules of different cities. Formulas recommended by the 
authors, and the requirements of several large cities, are given below. 


JOINT CoMMITTEE, 1924, RuLES ror STRUCTURAL STEEL AND Cast- 
IRON COLUMNS, INDORSED By AUTHORS 


The Joint Committee, 1924, provides rules for composite columns 
and structural steel columns. The allowable stresses for both types 


436 CONCRETE AND REINFORCED CONCRETE COLUMNS 


are the same, the difference being in the requirement as to details. 
These rules are indorsed by the authors. 


Let f’. = ultimate compressive strength of conerete * at 28 days, 

lb. per sq. in. 

f, = compressive unit stress in metal core, lb. per sq. in.; 

A = area of concrete core, sq. in.; 

A, = area of structural steel, sq. in. ; 

P = total safe axial load, lb.; 

R = least radius of gyration of the steel or cast-iron section, 
in. 

h = height of column, in. 


Composite Columns.—Composite columns are columns consisting 
of structural steel or cast iron encased in a spirally reinforced con- 
crete core. : 

The following formulas should be used for design of composite 
columns: 


Safe Load on Composite Column, 
P =(A — A,)fo +pAgie pe 


where 


Allowable Unit Stress in Concrete, 
f, = 0.25f". ys eee (42) 
Allowable Compressive Unit Stress on Steel Section, | 
f, = 18.000 — 705, a Se ee ema an 
but shall not exceed 16 000 Ib. per sq. in. 
Allowable Unit Stress on Cast Iron, 


J, = 12.000 — 605, Me ren rn ee 
but shall not exceed 10 000 lb. per sq. in. 

The diameter of the cast-iron section shall not exceed one-half 
of the diameter of the core within the spiral. The spiral reinforce- 
ment shall be not less than 0.5 per cent of the volume of the core 
within the spiral and shall conform in quality, spacing, and other 
requirements to the provisions for spiral columns. 


3 Ultimate strength based on tests of 6 by 12 in. or 8 by 16 in. cylinders. 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 437 


Ample section of concrete and continuity of reinforcement shall 
be provided at the junction with beams or girders. The area of the 
concrete between the spiral and the metal core shall be not less than 
that required to carry the total floor load of the story above, on the 
basis of a stress in the concrete of 0.35f’-, unless special brackets are 
arranged on the metal core to receive directly the beam or slab loads. 

Structural Steel Columns.—A structural steel column is a column 
consisting of a structural steel section which fully encases an area 
of concrete, and which is protected by an outside shell of concrete 
at least 3 in. thick. . 

Formulas for safe load and unit stresses are the same as for com- 
posite columns. 

The outside shell shall be reinforced by wire, mesh, ties, or spiral 
weighing not less than 0.2 lb. per sq. ft. of surface of the core, and 
with a maximum spacing of 6 in. between strands or hoops. Special 
brackets shall be used to receive the entire floor load at each story. 
The safe unit stress in steel columns, calculated by Formula (43), 
p. 436, shall not exceed 16,000 lb. per sq. in. 


RULES FOR STRUCTURAL STEEL or Various Burtpina CopEs 4 


New York Code Regulation for Structural Steel and Concrete.— 
In columns of structural steel, thoroughly encased in concrete not 
less than 4 inches thick and reinforced with not less than one per 
cent of steel, the allowable load shall be 16 000 lb. per sq. in. on the 
structural steel, the percentage of reinforcement being the volume of 
the reinforcing steel divided by the volume of the concrete enclosed 
by the reinforcing steel. Not more than one-half of the reinforcing 
steel shall be placed vertically. The reinforcing steel shall not be 
placed nearer than one inch to the structural steel or to the outer 
surface of the concrete. The ratio of length to least radius of gyra- 
tion of structural steel section shall not exceed 120. 


Dea gee GOOG cli te ee Bis hts Oo eee 


This regulation differs from the Joint Committee Rules in one 
essential, namely, that no direct strength is attributed to the con- 
crete, the increase being taken care of by increased unit stress in 
steel above that allowed for ordinary structural steel columns. 

Philadelphia Regulation for Structural Steel and Concrete,— 
The Philadelphia ruling follows closely the New York rule, except 


* Building Codes are given as of 1925. As changes occur they will be 
made in subsequent imprints of this treatise. 


4383 CONCRETE AND REINFORCED CONCRETE COLUMNS 


that the required sum of lateral longitudinal reinforcement outside 
of the shapes is to be 1 per cent of the concrete within the hoops, 
instead of within the steel section as required in the New York Code. 
Boston Rule.—No increase in strength is attributed to the 
‘mbedment of structural steel in concrete. It is permissible, how- 
ever, to utilize the concrete casing to carry the load, for one or more 
stories between the brackets on structural steel. The rule reads: 


“ Reinforced concrete buildings may be supported by structural 
steel or cast-iron columns, fireproofed in first-class construction as 
provided elsewhere in this act. Brackets shall be provided to trans- 
mit the load from the floors to the column. Such columns shall be 
computed as follows: 

“ (qa) If the brackets are placed immediately below the floor, the 
structural steel or cast-iron columns shall be assumed to carry the 
load of all the floors above. 

“(b) If the brackets are placed immediately above a. floor, the 
structural steel or cast-iron columns shall be assumed to carry all 
the load above the brackets and the floor or floors below the brackets 
shall be carried on reinforced concrete encasing the metal, designed 
in accordance with the requirements of this act, to the next bracket 
below or to the foundation. In this case, however, the surrounding 
concrete shall be so separated from the steel or cast iron as to permit 
the separate action of both. 

“ Circular hollow steel or wrought-iron columns filled with con- 
crete shall be allowed to carry a load equal to the capacity of the 
metal casing plus the capacity of the concrete filling. The average 
unit stress in the casing shall be that specified elsewhere in this act 
for columns, and that in the concrete filling shall be in the same ratio 
to the unit stress in the casing which the modulus of elasticity of the 
concrete bears to that of the casing.” 


This rule is not logical, as it entirely neglects the effect of the 
concrete surrounding the steel, although it acknowledges its strength 
by allowing it to carry the floor loads when the concrete is separated 
from the steel section. 

Cleveland Code.—The Cleveland Code divides the structural 
steel columns into two classes: the columns with solid web, and the 
columns with open web. For solid-web columns, no allowance is 
made for the strengthening effect of the concrete, on the assumption 
that no bond can exist between large flat surfaces of the steel section 
and the concrete. They must be designed with the same stresses 
as ordinary structural steel columns. For open-web columns, the 
unit stresses for the steel are increased over those allowed for struc- 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 439 


tural columns. The rate of increase depends upon the type of col- 
umn and also upon the ratio of the steel area to the area of con- 
crete. The larger the percentage, the smaller is the allowable 
increase in unit stresses. This last provision is logical. The con- 
crete section has a definite strength and rigidity. When a structural 
column is imbedded therein, the total strengthening effect of the 
concrete, within limits used in practice, will be practically the same, 
irrespective of the magnitude of the area of the steel column. The 
effect per unit area of the steel section will be larger if the steel sec- 
tion forms a small part of the total section and diminishes with the 
increase of the section. The rule reads as follows: 


* Structural Steel Columns Encased in Concrete.— 


“(a) Solid Web Columns.—In columns of this type, the stcel 
shall be designed to take the total dead and live loads, and the con- 
crete shall be considered as fireproofing only. All columns of this 
type shall be wrapped with wire in such a manner as to securely 
hold the concrete in position. All loose scale or rust shall be removed 
before encasing the column in concrete. 

“(b) Open Web Columns.—In columns of this type, the steel 
shall be designed to take the full dead and live loads with the follow- 
ing unit stresses, with no added allowance for the concrete: 


“Gray columns and similar types, 

20 000 — 300 X the percentage of‘steel. 
“ For four-angle columns, latticed four sides, 

19 000 — 300 X the percentage of steel. 
“For four angles with latticed web, 

17 500 — 300 X the percentage of steel. 


‘“Open-web columns shall be wrapped with one-eighth (4) inch 
(or larger) wire at vertical intervals not greater than eight (8) inches. 

“The percentage of steel shall be based on the total area of the 
column, after deducting two (2) inches all around. 

“If the unsupported length of columns exceeds twelve (12 
times the least outside dimension, the stresses shall be reduced as 
required for reinforced concrete columns. 

‘In all columns under this section, positive connections shall be 
provided to transmit to the column steel the loads of all rod rein- 
forced beams and girders framing into the column flanges.” 


Chicago Requirements.—The Chicago Code provides for three 
combinations of structural steel with concrete. 
First.—‘‘Structural steel of box shape with lattice or batten plates 


filled with concrete. If no structural shape is less than 1 sq. in. in 
section and the spacing of the lacing or battens is not greater than 


440 CONCRETE AND REINFORCED CONCRETE COLUMNS 


the least width of the columns, and the structural steel does not 
exceed 8 per cent of the area, then the column may be considered 
as a reinforced concrete column and designed by regular formulas 
in which the stress in concrete equals one-quarter (4) of the ultimate 
strength of concrete and the unit stress in steel equals the unit stress 
in concrete multiplied by ratio n.”’ 

Second.— For steel columns filled with, and encased in concrete 
extending at least three inches beyond the outer edge of the steel, 
where the steel is calculated to carry the entire live and dead load, 
the allowable stress per square inch shall be determined by the 
following formula: 


18 000.--.70-2, 1b.,°. sea 


but shall not exceed 16 000 pounds.” | 
Third.—“ For steel columns filled with, but not encased in con- 
crete, the steel shall be calculated to carry the entire live and dead 
load. In this case, the above formula may be used, but the allowable 
stresses shall not exceed 14 000 pounds.” 
The above stresses should be compared with the following stresses 
allowed for structural steel columns not encased in concrete. 


16 000 — 705 lb. persqy inst 2) a eaee eae 


In the above formula 


L = length of column in inches; 
R = least radius of gyration in inches. 


Deraits ofr DESIGN OF STEEL SECTION 


The details of the steel section depend largely upon conditions 
and upon the building department requirements. Naturally, when 
no allowance is made for the effect of the concrete surrounding the 
steel, the main object will be to get the best and cheapest steel sec- 
tion. On the other hand, when concrete is relied upon to resist 
stresses, it is desirable to get cooperation between the two materials, 
and a section must be selected which will provide best bond between 
the concrete and steel. The simplicity of the brackets and splices 
should also be taken into consideration, and a section selected which 
requires the least complicated brackets at the floor levels to transmit 
the floor load to the column. . 

Figure 149, a to f, inclusive, give the various sections in use. 
The cheapest section is likely to be the one requiring least shopwork 
and the one in which no material is lost for stay plates and lattice 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 441 


work. Such is the section shown in Fig. 149, e. The objection 
to it is that it cannot be counted upon to cooperate with the con- 
erete. In using this section, it is important to bind the concrete 
behind the flanges by means of ties, as otherwise the thin concrete 
may spall off. This section is accepted by the City of New York and 
by Philadelphia. In Cleveland, no increase of strength is allowed 
on such section. It can be used in connection with the Boston law, 
except that when concrete is depended upon to carry the floor load 
from floor to floor, the concrete behind the flanges cannot be con- 
sidered, as it is too thin. 












Y7 LLikifg lls 
- 


Tie 
or Spiral” 





_-Tie 






~. Tie 
or Spiral 





Fig. 149.—Structural Steel Columns Encased in Concrere: (See p. 440.) 


The Gray Column, Fig. 149, b, is the best type of a concrete 
column. ‘The objection to it is the large cost of fabrication. 

The Star Shape Column, Fig. 149, d, has been used to some 
extent, as it is compact and fits well into a round column. It is par- 
ticularly adaptable under the Boston Law. It should not be used 
where considerable bending is expected unless bending moment is 
resisted by vertical bars. 

Splicing Structural Columns.—Splices should be designed in 
the same manner as for steel columns. No reliance should be placed 
on the concrete. The splices are best placed above the floor level. 


442 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The design of splices will depend upon whether the ends are 
machined or not. 

If the ends are not machined, to insure good bearing on the 
lower section, the splice must be strong enough to transfer the full 
load through the rivets to the lower column. ‘The number of rivets 
must be large enough to transfer the load from the column to the 
splice and then from the splice to the lower section. , 

If the ends are machined, as is generally the case, the load from 
the upper column is transferred directly by bearing to the section of 
the lower column. Only enough strength is required in the splice 
to keep the section securely in place during construction and to 
resist any bending moment to which the column may be subjected. 
In some types of columns, the concrete section may be counted upon 
to resist the bending moment, in connection with a proper amount 
of bars used to resist any possible tension. 

Steel Brackets.—The construction of steel brackets to transmit 
the floor load to the steel column depends upon the cross section of 
the steel column and also upon the construction of the floor. 

With a closed section, such as Fig. 149, d, ¢, or f, it is neces- 
sary to provide brackets at every floor for the load coming on the 
flange sides. In other sections, brackets may be placed only every 
two or three floors and concrete relied upon to transmit the load partly 
by bond to the steel section and partly to the brackets. In such 
cases, of course, the concrete must be strong enough to carry the 
floor loads as a column between the brackets. Under no circum- - 
stances should the brackets be omitted altogether and the bond relied 
upon to transfer the entire load. 

In flat slab construction, the brackets are usually placed within 
the column head. Angles properly riveted are sufficient. For 
heavy loads, stiffeners under the angles may be required. The 
angles should be placed in such a position that they will not inter- 
fere with the pouring of the concrete. This fixes their position away 
inside of the column head where the width of the column head form is 
much larger than the column, so that there is room for the concrete 
to flow into the column form. In concreting, care should be taken 
that the column is properly filled with concrete, especially when it 1s 
counted upon, directly or indirectly, to strengthen the column. 

In beam and girder construction, brackets should be supplied 
for every girder. These should be strong enough to transmit all the 
load to the steel section. When concrete is relied upon to transmit 


STRUCTURAL STEEL COLUMNS IMBEDDED IN CONCRETE 443 


the load to the story below, it should be of proper dimensions and 
strength to carry the load as a column. 
Examples of brackets are shown in Figs. 150 and 151. 





Fig. 150.—Steel Brackets for Floor Beams. (See p. 442.) 





4-33 
ZZ pana 
RAS 


Fig. 151.—Steel Brackets for Flat Slab. (See p. 442.) 


Details of Base for the Steel Section.—The load from the steel 
section is transmitted to the concrete of the foundation by means 
of a base designed in the same manner as for structural steel columns, 


444 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The base for a structural column is either cast or built up of plates 
and angles. Where easily obtainable, a base of thick solid slabs of 
rolled steel may be used. The thickness of the plate may run up to 
6 in. 

The problem in designing the base is to spread the load from the 
column on a sufficiently large area of the concrete foundation, so as 
not to exceed the allowable bearing stresses on concrete. After the 
area of the base is determined by dividing the column load by the 
allowable bearing stress, the base plate is strengthened, by angles in 
built-up bases, and in cast-iron bases by ribs of sufficient proportions 
to resist the stresses produced by the upward pressure of the foun- 
dation. 

The thickness of solid rolled plate is determined by the bending 
moment due to the upward reaction. It may be obtained from the 
following formula. 


| 


Let P = column load, lb.; 
a and b = dimensions of foundation plate, in.; 
c and d = outside dimensions of column, in.; 
t = thickness of plate. 
Then, for a stress in steel of 16 000 Ib. per sq. in., 
Thickness of Foundation Plate, 


(48) 


ata alti 3, P(b — d) 
t= 4] 213006 q 


21 300a ~ 










Cast Iron 
wr 


Fig, 152.—Bases for Steel Columns. (See p. 443.) 


For square plates, both dimensions are equal. For rectangular 
plates, the larger of the two values should be accepted. 


ECONOMIES IN COLUMN DESIGN 445 


The column must be bolted or riveted to the base. The bases 
must also be bolted to the foundation. If uplift is possible, the bolt 
must be strong enough to resist it. The bolts should be carried so 
far into the concrete, and anchored in such a way, that enough weight 
of concrete above such anchorage is 
available to counteract the uplift. 

Figure (a), (b), and (c) shows bases 
adaptable for steel colums. 

Cast-iron Core Imbedded in Con- 
crete.—A section of a column con- 
sisting of a cast-iron core imbedded 
in concrete is shown in Fig. 153. 
The cast-iron core is spliced by means 
of a pipe sleeve fitted on the outside Fic. 153.—Details of Column with 
of the core. Where the diameter of ©28t-iron Core. (See p. 445.) 
the core changes, special sleeve is 
provided. The ends of the core must be faced to even bearing 
on the lower core, 


_- Cast Iron 






. _Vertical Bars 


\---Spiral 


ECONOMIES IN COLUMN DESIGN 


Economical Dimensions and Concrete Proportions.—In a one- 
story structure, the problem is to select a column for which the cost 
of materials is a minimum. For structures several stories high, the 
additional question of formwork is introduced into the problem. 
The column load varies from story to story. If the cost of materials 
alone were considered, the greatest economy would be obtained by 
changing the size of the column at every story. However, the reduc- 
tion in column size requires remaking of column forms (except where 
they are easily adjustable) and also the lengthening of forms for beams 
and girders running into the column. Changing of column size is 
economical when the cost of these changes in formwork is smaller 
than the saving in materials resulting from the reduction in the size 
of the column. In many cases, the changes in formwork cost more 
than the saving in materials. No general rule can be established, 
since the relation between the cost of materials and that of labor 
varies greatly in different localities. As a general proposition, it is 
safe to say that the columns should be carried the same size for at 
least two stories. In many instances, the same size of column may 


446 CONCRETE AND REINFORCED CONCRETE COLUMNS 


be maintained through several stories, by successively reducing the 
amount of reinforcement and changing the mix of concrete. 

The above suggestions do not apply to round columns, especially 
in flat slab construction. The column and column-head forms in 
such cases are usually made of sheet metal. As they are readily 
adjustable, no saving in cost is obtained by maintaining the size of 
the column unchanged. It should be remembered, however, that 
for standard forms the diameter of column is adjustable only for 
even inches; therefore, the diameter of column should always be in 
even inches. 

To get best results, it is desirable that the designer should become 
acquainted with cost data in the location in which the job is to be 
built. 

It is good practice to use the same mix of concrete for all columns 
on one floor. An exception may be made in case of wall columns, for 
all of which leaner mix may be used than in all interior columns. 
Confusion may result if some of the interior columns are made of one 
mix and the rest of some other mix. » 

Wall columns are usually rectangular in shape. To reduce form- 
work, it is advisable to maintain the same width of the column through 
its whole height and to effect the change in cross section by changing 
the thickness of the column. ‘This method reduces the cost of remak- 
ing formwork, as all remaking is confined to two sides; also the 
forms for spandrels are used without remaking. This suggestion 
does not hold in cases where, for architectural considerations, special 
designs of column are required. In such cases, the shape of the 
concrete column must conform to the shape of the finished pier. 

Where possible, the concrete column should extend to the edge of 
the window sash, and recesses should be provided in the concrete to 
‘receive the sash. The width of the column should then be fixed 
exactly after the width of the sash is determined. Very often, this 
width is in fraction of an inch as it is obtained by subtracting from 
the span the fixed width of the sash. 

It is desirable to use columns of the same diameter throughout 
each floor of the building. When the loads of some odd columns in a 
floor are smaller than those of the typical column, but the difference 
is not appreciable, it may be advantageous to use not only the same 
concrete size, but also the same amount of steel as for the typical 
column, especially when it is difficult to distinguish the odd columns, 
by any peculiarity in their location, from typical columns. If a 


ECONOMIES IN COLUMN DESIGN 447 


a smaller amount of steel is used for odd columns there is danger of 
interchanging the special reinforcement with the reinforcement of 
the typical column, to the detriment of the latter. Of course, this 
does not apply to special columns, such as columns in elevator walls 
and staircases, which, because of their peculiar locations, are not 
likely to be confused with the typical columns. 

Economical Mix of Concrete.—Under average conditions, the 
cost of 1 : 1 : 2 mix of concrete in place ° is about 14 per cent greater 
than that of the 1: 14:3 mix, while the strength of 1:1: 2 is 
18 to 21 per cent greater than that of the 1: 15:3. The cost of the 
1:1: 2 is 24 per cent greater than that of the 1 : 2: 4 mix, and the 
strength of the 1: 1: 2 is 45 to 50 per cent greater than that of the 
1:2:4. The cost of the 1: 15:3 mix is about 8 per cent greater 
than that of the 1 : 2: 4 mix, and the strength is 20 to 27 per cent 
greater. : 

From comparison of the increase in cost of richer mixture with 
the increase in allowable stresses, it is evident that the percentage of 
increase in strength is much larger than the percentage of increase in 
cost. Therefore, the richer mix is more economical for columns 
than the leaner mix. To obtain best economy, therefore, a rich mix 
of concrete should be used, except in top columns where the size 
may be governed by the requirements as to minimum size and 
1:2:4 concrete gives sufficient strength. For lower columns, 
- instead of increasing the size of the column, richer mix should be used. 

Economical Design of Columns with Longitudinal Reinforce- 
ment.—The cost of steel in a column is forty to fifty times larger 
than the cost of concrete of the same volume, depending upon the 
mix of concrete. The smaller value is for 1: 2:4 mix and the 
larger value for 1: 1:2 mix. The allowable stress on vertical bars 
is only from ten to fifteen times larger than the allowable stress on 
concrete. It follows that it is much cheaper to use concrete than 
steel to support the load. Therefore, if it is required to increase the 
carrying capacity of a column, it is cheaper, so far as the cost of 
materials is concerned to increase the size of the column than to 
add vertical steel. 

The most economical column, then, is one in which the minimum 


> The cost of concrete used in comparison includes cost of materials, cost of 
plant, and cost of mixing and placing. Cost of formwork is not included. Richer 
concrete requires smaller columns. In some cases, this reduces the cost of form- 
work, thus increasing the advantage of rich mix. 


448 CONCRETE AND REINFORCED CONCRETE COLUMNS 


permissible amount of vertical steel is used. Some building codes 
allow 4 per cent of vertical bars as a minimum. ‘The authors do 
not recommend the use of a smaller amount of vertical steel than 
1 per cent of the effective area of the column. Exception may be 
made to this rule in case of wall columns in which the column serves 
as a pilaster, and its area, as determined by architectural require- 
ments, is larger than required by the stresses. In such cases, the 
area of column required by the load for the specified stresses and for 
1 per cent of steel should be computed and the area of steel to be 
used should be made equal to 1 per cent of the area of column as 
determined. (See example 4, p. 410.) The steel so obtained should 
be distributed over the whole section and the proper amount of ties 
used. If the column is also subjected to bending, proper bending 
reinforcement should be provided. | 

Economical Design of Spiral Columns.—The economical design 
of spiral columns will differ for the different types of formulas now 
in use. The economy of columns designed according to the formulas 
given on pp. 421 to 426 is discussed below. : 

Columns with Fixed Percentage of Spiral—Under this head 
come the columns designed according to the authors’ reeommenda- 
tions and 1916 Joint Committee recommendations and according to 
Boston, Cleveland, and Philadelphia Codes. The formulas used 
in design of spiral columns are the same as for columns with vertical 
steel only, with the exception that larger unit stresses are allowed. 
The relation between the ratio of cost of steel to cost of concrete and 
the ratio of unit stresses is same as for columns with vertical steel 
only. As in the case of columns with bars only, the most economical 
spiral column is one with minimum allowable amount of vertical 
steel. Rich mix of concrete also is more economical than lean mix, 
as the allowable strength increases more than the cost of concrete. 

Columns with Variable Percentage of Spirals.—Under this head 
come the rules of Chicago and New York. 

Chicago Code-—Formula (20), p. 425 [P=Af.-+ n—- Ase 
+ 2inpiAf-] has three variables: area of concrete, percentage of 
vertical bars, and percentage of spirals. The effectiveness of vertical 
bars is the same as for columns with vertical bars only. They are, 
therefore, equally uneconomical means of increasing the strength in 
both cases. The spiral reinforcement in the formula is considered 
as two and one-half times as effective as vertical bars of equal volume. 
It is, therefore, a much more economical means of increasing the 


ECONOMIES IN COLUMN DESIGN A449 


strength of the column than the vertical bars. Comparing with 
concrete, the increase in strength produced by the spiral is 2.5n 
_ times’as large as that produced by the same amount of concrete. 
This ratio equals 25 for rich concrete and n = 10, and 37.5 for lean 
concrete and n= 15. The ratio of the cost of spiral to that of 
concrete of equal volume is almost always larger than the ratio 
of increase in strength, consequently, it is cheaper to increase the 
strength of column by increasing its size than by increasing the per- 
centage of spiral, especially as an increase in the percentage of spiral 
reinforcement requires an equal increase in vertical steel. The most 
economical spiral column designed in accordance with the Chicago 
Code, then, is the one which has the smallest amount of spiral and 
vertical steel, but in which the concrete is of the richest mix. 

When the size of column is fixed, the cheapest means of increas- 
ing its strength is by an increase in the amount of spiral reinforce- 
ment and a proportional increase of vertical steel, rather than by an 
increase in the amount of vertical steel alone. 

New York Code——In Formula (16), p. 424, for-spiral columns ac- 
cording to this code [P = Af, + (n — 1) Asfe-+ 2p, Af,] the effect of the 
spiral reinforcement is sixty-seven times larger than that of 1: 11:3 
concrete of the same volume and eighty times larger than that of 
1: 2:4 concrete of the same volume.® Since the ratio of the cost 
of steel to the cost of concrete is usually smaller than the above 
values, it is evident that the spiral reinforcement is the most eco- 
nomical material in the make-up of the column so designed. The 
vertical steel, as in the previous cases, is the least economical material. 
The most economical column, then, is one in which the percentage 
of the spiral is the maximum allowable and the amount of vertical 
steel the least allowable. The richer mix is also more economical 
than the leaner mix for the reasons given in connection with col- 
umns with vertical bars. Although there is no requirement fixing 
the ratio of the percentage of vertical steel to that of the spiral, for 
good design it is advisable to use at least as large a percentage of 
vertical bars as of spirals. 

Relative Economy of Spiral Columns and Columns with Vertical 
Steel Only.—The relative economy of the two types of columns 
depends upon the formulas used in designing. As a general proposi- 


2 
6 This ratio is obtained from = using fs = 20 000 Ib. per sq. in. and te = 500 
Cc 


and 600 Ib. per sq. in. respectively. 


450 CONCRETE AND REINFORCED CONCRETE COLUMNS 


tion, spiral columns are more economical than columns with vertical 
steel only, with the exception of columns designed by the New York 
Code, as explained below. 

Boston Cleveland, and Philadelphia Codes.—In columns designed 
in accordance with these codes, spiral amounting to 1 per cent of 
the effective area is used, and the effect of the spiral is taken care of 
by an increase in the allowable unit stresses on concrete. Referring 
to tables on p. 409 and on p. 423, this increase amounts to more than 
50 per cent of the stresses used for columns without spiral reinforce- 
ment. 

For instance, the allowable stress for 1: 2:4 concrete, according 
to the Boston Code, is 495 lb. per sq. in., while for spiral columns 
the allowable stress is 770 lb. per sq. in. The increase in strength 
amounts to 275 lb., which is 56 per cent of the lower value. In addi- 
tion, the unit stress in vertical steel is increased in the same ratio. 

To get the relative economy of a spiral column and of a column 
with vertical bars only, it is necessary to determine the increase in 
cost produced by the use of the spiral. This depends upon the ratio 
of the cost of spiral to the cost of concrete. Under present conditions 
(year 1925), the ratio of the cost of spiral to the cost of equal volume 
of concrete is 45 for 1 : 2: 4 concrete, A1.5 for 1: 14: 3 concrete, 
and! 36.for te: li 22 ‘contrete ier these unit costs, 1 per cent of 
spiral increases the cost of column by 36 to 45 per cent of the cost 
of the concrete within the effective area. 

Comparing the imcrease in allowable stresses (56 per cent) with 
the increase in cost (386 to 45 per cent), it-is evident that the increase 
in strength is much larger than the increase in cost; consequently, a 
spiral column is more economical than a column with vertical bars 
only. It may also be stated, on the basis of the above and of the 
discussion on p. 447, that the most economical column designed 
according, to the above codes is a spiral column of richest mix having 
the minimum allowable percentage of steel. 

New York Code.—According to the New York Code, the spiral 
reinforcement does not affect the unit stresses in concrete. It 
increases the strength of the column directly, the increase being 
expressed by the term Qi Afs (see p. 424). As explained on p. 449, 


the ratio of increase in strength effected by the spiral reinforcement is 


7 Based on cost of spiral in place of 41 cents per lb. and on following costs of 
concrete per cu. ft. in place: 1:2:4 mix, 50 cents; 1: 14:3 mix, 54 cents; 
and 1:1: 2 mix, 62 cents. 


ECONOMIES IN COLUMN DESIGN 451 


larger than the increase in cost. The deduction would be that 
spiral columns are more economical than columns with vertical bars 
only. This is not always true, however, because in computing the 
strength of a column with vertical bars only, the Code does not 
require any allowance for fireproofing, but permits the use of the total 
cross section of the column as effective area. In spiral columns, on 
the other hand, 2-in. fireproofing is required, so that for equal diam- 
eter of column the effective area of a spiral column is less than the 
effective area of a column with vertical bars only. The difference is 
particularly large in columns of small diameter. For this reason, 
spiral columns designed as explained above are not economical for small 
diameters of columns although they are economical for large diameters. 

Chicago Code.—According to the Chicago Code, spiral reinforce- 
ment increases the strength of column directly, as expressed by the 
term 23npiAf,, and indirectly by increasing the allowable unit stresses. 

The direct increase in strength due to the spiral equals twenty-five 
times the strength of an equal volume of concrete for 1: 1 : 2 mix; 
ov times for 1:13:38 mix; and 374 times for 1:2:4 mix. 
This means, for 1 per cent of spiral, an increase in the average strength 
in concrete over the whole effective area amounting to 25 to 373 
per cent above the allowable stress on concrete for spiral columns, 
and 31 to 47 per cent above the allowable stress on columns with 
vertical steel only. The indirect increase in strength is caused by 
the larger allowable unit stress, f-, on spiral columns. 

While, for columns with vertical steel only, the allowable unit 
stress equals 3 of the ultimate compressive strength, the working 
stress in spiral columns amounts to } of the ultimate strength. 
This means an increase in working stresses, due to the spiral rein- 
forcement, of 25 per cent, not only for the concrete, but also for 
the vertical steel. 

When the direct and the indirect increases in strength, produced 
by the spiral, are added, the total increase in strength of a spiral 
column over a column with vertical steel only is found to be from 
56 to 72 per cent, depending upon the mix of concrete. 

The increase in cost due to 1 per cent of spiral amounts to 36 to 
45 per cent of the cost of concrete, as explained on p. 449. 

Comparing the increase in cost of concrete with the increase in 
strength of column, it is evident that the increase in strength is 
larger than the increase in cost. Therefore, a spiral column is more 
economical than a column with vertical steel only. 


452 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Economy of Structural Steel Columns Imbedded in Concrete.— 
The use of structural steel in columns is not economical. It should 
be restricted to cases where a reduction in size of column is required, 
beyond that possible with spiral columns of rich mix and with the 
largest. possible percentage of vertical steel. Usually, the struc- 
tural steel is used in the lowest floors only, while in upper floors 
concrete columns are substituted as soon as the reduction in load 
permits it. 

The objections to structural steel columns are as follows: 

Greater Cost of Structural Steel Columns as Compared with Rein- 
forced Concrete Columns.—Structural steel columns imbedded in 
concrete always cost more than reinforced concrete columns eco- 
nomically designed to carry the same load. ‘The increase in cost 
depenYls upon the specification used. It 1s, of course, much larger 
for specifications in which no allowance is made for the strengthening 
effect of the concrete. With the formulas recommended by the 
authors, the increase in cost is considerably reduced, as both steel 
and concrete are utilized in carrying the load. | 

Difficulties in Transferring the Floor Load and Tying the Floor to 
the Column.—These are particularly serious in the case of beam and 
girder construction, where it may be difficult to make the beams and 
girders continuous. In flat slab construction, the difficulties are 
considerably reduced, especially with systems of flat slab reinforce- 
ment in which a small number of bars, or none at all, cross the 
column (see p. 369). 

Difficulties in Erecting Steel Columns.—The steel cores are very 
heavy, usually weighing more than a ton. This offers a special 
problem in erection in a reinforced concrete job not equipped with 
facilities for lifting heavy members in the manner used in structural 
steel buildings. Therefore, it is either necessary to introduce special 
equipment on the job for handling the steel cores or to handle them 
at large expense without proper equipment. In either case, it is 
costly. To reduce the weight, it is advisable to limit the height of 
the steel cores to one story. 


REDUCTION OF LIVE LOAD IN BUILDINGS 


In designing columns, all the dead load and a certain proportion 
of the live load shall be used. By dead load is meant the weight 
of the floor system, floor finish, fixed partitions or walls, columns, 


REDUCTION OF LOVE LOAD IN BUILDINGS 453 


and other fixed loads. Live load is the load that comes on the floor 
in the course of its use. It is generally recognized that a larger live 
load must be allowed for the floor system than for columns, because 
overloading of a small section of the building is much more likely to 
occur than simultaneous overloading of a number of floors carried by a 
column. The likelihood of the latter occurrence decreases with the 
increase in the number of floors to be supported. To allow for this, 
the reduction is small for the columns in the top floors and increases 
with the number of floors. 

Authors’ Recommendation.—The authors recommend that in 
all buildings the columns be designed for the total dead load, plus a 
proportion of total live load above the columns given in the table 
below. In this table are given proportions to be used for ordinary 
buildings and also larger proportions for warehouses and ReaNY. 
manufacturing buildings. 


Recommended Proportion of Live Load on Columns 





Proportion of Live Load 
on Columns 


Number of Floors 


Carried by 
Seine ney, Warehouses 
Buildings 
1 1.00 1.00 
2 0.85 0.90 
3 0.80 0.87 
4 0.75 0.84 
5 0.70 0.81 
6 0.65 0.78 
7 0.60 0.75 
8 0.55 . 0.75 
9 0.50 0.75 
10 0.50 0.75 


For use of this table see Example 6 on page 456. 

Usually, no reduction is allowed for warehouses, the idea being 
that in a warehouse all floors are apt to be fully loaded at the same 
time. However, the chances of overloading a column in a ware- 
house are unquestionably much smaller than the chances of over- 


A454. CONCRETE AND REINFORCED CONCRETE COLUMNS 


loading a floor panel. In all warehouses, moreover, free aisles must 
be left for moving the goods. These two considerations convince 
the authors that some reduction is permissible for warehouses. 
The reduction recommended is smaller than for ordinary buildings. 


Boston Code.—The Boston Code makes the following provisions 
for reduction of live load: 


In all buildings except storage buildings, wholesale stores, and 
public garages, for all columns, girders, trusses, walls, piers, and 
foundations. 


Carrying one floor........ ashe kites wien ee No reduction. 
Carrying two floors..........+seeeeeeeees 25 per cent reduction. 
Carrying three floors..........++++++++++: 40 per cent reduction. 
‘Carrying four floors2+). 7.325264 ao aaeae 50 per cent reduction. 
Carrying’ five floors mee .4 sae ote 55 per cent reduction. 
Carrying six floors or more........--.---- 60 per cent reduction. 


In public garages, for all flat slabs of over three hundred square 
feet area reinforced in more than one direction, and for all floor 
beams, girders and trusses carrying over three hundred square feet 
of floor, and for all columns, walls, piers, and foundations, twenty- 
five per cent reduction. 


New York Code.—New York allows a reduction of live loads only 
for buildings over five stories high. ‘The Code states: 


Loads on Vertical Supports—Every column, post, or other verti- 
cal support shall be of sufficient strength to bear safely the combined 
live and dead loads of such portions of each and every floor as depend 
upon it for support, except that in buildings more than five stories 
in height the live load on the floor next below the top floor may be 
assumed at ninety-five per cent of the allowable live load, on the 
next lower floor at ninety per cent, and on each succeeding lower 
floor at correspondingly decreasing percentages, provided that in no 
case shall less than fifty per cent of the allowable live load be 
assumed. 


Chicago Code.—The Chicago Code gives the following rules for 
reduction of live loads on columns: 


Walls, Piers and Columns—Dead and Live Loads.—(a) The full 
live load on roofs of all buildings shall be taken on walls, piers, and 
columns. 


REDUCTION OF LIVE LOAD IN BUILDINGS 


455 


(b) The walls, piers, and columns of all buildings shall be designed 
to carry the full dead loads and not less than the proportion of the 
live load given in the following table: 


Floor 17 | 16/15/14) 13) 12) 11/10 











% 
17 85] % 
16 60! 85] % 
15 75| 80| 85] % 
14 70) 75| 80| 85! % 
13 65| 70] 75] 80| 85 
12 60| 65] 70} 75] 80 
11 55| 60] 65] 70) 75 
10 50} 55] 60| 65| 70 
9 50! 50] 55) 60] 65 
8 50] 50! 50! 55] 60 
7 50| 50) 50) 50) 55 
6 50} 50) 50) 50| 50 
5 50) 50] 50] 50] 50 
4 50| 50) 50) 50) 50 
3 50) 50} 50] 50) 50 
2 50) 50) 50] 50] 50 
1 50) 50) 50} 50| 50 











Zo 
85 
80 
75 
70 
65 
60 
55 
50 
50 
50 
50 





Zo 
85 
80 
75 
70 
65 
60 
55 
50 
50 
50 





Griese ere Golo, pete Bee 2! ae 


SS) SS a ee SS 


To 
85 


80 
75 
70 
65 
60 
55 
50 
50 


Zo 
85 
80 
75 
70 
65 
60 
55 
50 


To 
85 
80 
75 
70 
65 
60 
55 


To 
85 
80 
75 
70 
65 
60 








Jo 


85| % 
80} 85} % 
75| 80} 85 


(c) The proportion of the live load on walls, piers, and columns 
on buildings more than seventeen stories in height shall be taken in 


same ratio as the above table. 


(d) The entire dead load and the percentage of live load on base- 
ment columns, piers, and walls shall be taken in determining the 


stress in foundations. 


(c) In addition to the entire dead loads, not less than the follow- 
ing proportion of the percentage of live load on the basement columns, 
piers, and walls shall be taken in determining the number of piles for 


pile foundations and the area of concrete caissons. 


Classes I and VII 
Classes IT, III and VI 


eee ete eee eee ee eee eee ee ee ee we eee 


ocee eet ee eee eee eee ee ee ee ee we 


amy Mv fanideV LEL! Aire Ml SE el el BA Pg 


In all foundations eccentric loading must be provided for. 


75 per cent. 
50 per cent. 
25 per cent. 


456 CONCRETE AND REINFORCED CONCRETE COLUMNS 


DESIGN OF COLUMN IN A BUILDING SEVERAL STORIES HIGH 


The procedure in designing columns in a building several stories 
high is given in the example below. 


Example 6.——Design an interior column in a building with ‘six stories and 
basement, where the conditions are as follows: Panel, 20 ft. square; Live loads, 
1st and 2nd floors, 250 lb. per sq. ft.; 3rd to 6th floors, 150 lb. per sq. ft.; Roof, 
40 lb. per sq. ft. 


Concrete dimensions are: 


1Istand 2nd Floor 3rd to 6th Floor Roof 


Slabs ceca t pate 4 in. 7k in. 6 in. 
Drop panel,.....°. 4 in. og in. 3. in. 


Dimension of drop panel, 7 ft. 6 in. square. 

Column head, 4 ft. 6 in. 

Story heights, 12 ft. from floor level to floor level. 

Floor finish consists of one inch granolithic applied separately. 

Stresses recommended by the authors on pages 407 and 421 will be used. 
Reduction of live load recommended for warehouses on page 453 will be used. 


Solution.—The loads per panel in each floor will be computed first. Live 
load and dead load will be kept separately. 


Area of panel, 20 X 20 = 400 sq. ft. 


Ist and 2nd Floor 3rd to 6th Floor Roof 
Live load......... 250X400=100kips 150X400=60kips 40400=16 kips 
Sigh load i 2eue ce. 106 100 75 
Finish vias sie 12 be 18 
Witt ae eee ee 118X 400 =47 112 400 =45 93 X 400 = 37 
Drop panel........ 3 2 2 
Total dead load... . 50 kips 47 kips 39 kips 


Kips is an abbreviation of ‘‘ Kilo pounds ” and means thousands of pounds. 

Further work will be simplified by tabulating the values as shown in table, 
page 457. After the design load is computed the size of columns is adopted in 
accordance with suggestions contained in this chapter. The required amount of 
reinforcement is finally found. The complete design of the column is given in 
the table. 





457 


DESIGN OF A COLUMN IN A BUILDING 


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Cogr ‘d ay) NOISAGC NWNIOO ALAAMONOD GAOAOANIAA AO AIA WVXA 


458 CONCRETE AND REINFORCED CONCRETE COLUMNS 


The table may be simplified by eliminating the items ‘“‘ Columns 
and Column Head.” This could be done by adding to the dead load 
of the floor the average weight of the column and column head. 

In case where no live load reduction is permitted and no division 
of live and dead load is desired for designing footings, the live and 
dead load may be combined. The first column in the table, then, 
gives the design load. 

For wall columns the weight of the curtain wall may be added 
to the slab load if it is constant in all floors, or treated as a separate 
item. 


COLUMNS SUBJECTED TO BENDING 


Bending in Interior Columns.—For equal spans and uniform 
loading, there will be no bending in the interior columns for the dead 
load or for the_live load extending over all spans. Some bending 
moment will be developed when the panels on one side of the column 
are loaded and those on the other side are not loaded. Usually, no 
allowance is made for the bending stresses thus produced. The 
bending moment in a column due to unbalanced live load may be 
determined as given in Volume III. 

When the spans are not equal, bending moments will be developed 
+n the interior columns both by dead and live loads. The bending 
stresses produced by this bending may be considerable and should 
be properly provided for. . 

Bending moments should be computed by formulas given in 
Volume III. ‘These should be combined with the axial load in the 
manner explained in connection with wall columns. If necessary, 
tension reinforcement should be provided. This is required at the 
top of the column near the face outside the largest span, and at the 
bottom near the face inside the largest span. 

Wall Columns.—When wall columns are connected by reinforce- 
ment with the floor construction, negative bending moment is devel- 
oped in the floor construction at the column. This bending moment 
produces bending stresses in the wall column. The wall column, 
then, is subjected to the direct compressive stresses produced by the 
superimposed loads and to the bending moment transferred from 
the floor to the column.® 


8 See ‘Design of Wall Columns and End Beams” by Edward Smulski, Journal 
American Concrete Institute, July, 1910. 


COLUMNS SUBJECTED TO BENDING 459 


The magnitude of the bending moment at the wall column for 
ordinary conditions is given on p. 280. For special conditions, the 
rules given in Volume III should be used. 

In some cases, additonal bending moments in the wall column 
are developed by eccentrically applied spandrel loads or by the 
weight of the cornice. These bending moments may be of the same 
sign as the bending moment transferred from the floor construction, 
in which case they should be added to it; or they may be of opposite 
sign, in which case they should be subtracted from it. 

In the top floor, the bending moment from the roof must be 
resisted by the top column alone. In the lower floors, part of the 
bending moment is resisted by the column below and part by the 
column above. The proportion of bending moment resisted by each 
of the two columns depends upon their rigidity, as expressed by 
: and a where J and J; are moments of inertia and h and h, are 
heights of the respective columns. If these ratios are equal or nearly 
equal, each column may be assumed as resisting one-half of the bend- 
ing moment from the floor. 

If the difference between the ratios of rigidity is appreciable, the 
proportion of the bending moment, /, resisted by the column having 








a ratio of stiffness equal to h is 7 i 7 M. For the other column, 
1 
hy 
I, 
the bending moment equals 7 hi 7 M. The sum of the bending 
non 


moments is M. 7 

It should be noticed that the moment in the column below the 
floor is negative, i.e., producing tension on the outside face of the 
column; while the moment above the floor is positive, producing 
tension at the inner face of the column. The deflection of the 
wall column is shown in Fig. 154, p. 460. The points of maximum 
tension are indicated there. 

Points of Application of Maximum Bending Moment.—The 
maximum negative bending moment may be considered as applied 
at the bottom of the beam or girder of the floor construction which 
produces the bending, and the maximum positive bending moment 


460 CONCRETE AND REINFORCED CONCRETE COLUMNS 


directly above the floor. 


Recommendation for flat slabs is given 


on p. 315. If the respective bending moments are plotted at the 
points of application of maximum moments, the straight line connect- 


_~ Beam before Bending 
4 Tension. 
SE Ie hn, BER, a oa 










‘Tension 


~~Beam after Bending 
Column before Bending 
“Column after Bending 


4~<-Tension 


Tension, 


Tension 


rhe ee 


\ _...Tension 


| “~~. Beam after Bending 
“---Qolumn before Bending 

eee 

\ ~--Column after Bending 


--Tension 





Fic. 154.—Deflection of End Beams and Wall Columns. 


(See p. 459.) 


ing them gives the 
variation of bend- 
ing moments in the 
column. 

Steps in Design 
of Exterior Col- 
umns.—After the 
column loads and 
bending moments 
are determined, it 
is necessary to com- 
pute the maximum — 
compression stresses 
for the combination 
of the concentric 
vertical column load | 
and the bending 
moment, and to de- 
termine the amount 
of tensile reinforce- 
ment required to 
resist the tensile 
stresses for the most 
unfavorable combi- 
nation of the con- 
centric column load 
and the bending 
moment. 

The first step 
is to determine the 
required dimensions 
for the column for 
the concentric col- 
umn load alone. 

Next, the maxi- — 


mum compression stresses are computed for the combination of 
maximum concentric column load and the maximum bending 


COLUMNS SUBJECTED TO BENDING 461 


moment. Hither formula on p. 177, or formulas on p. 182, may be 
used, depending upon the amount of tensile stresses developed for 
this condition of loading. The maximum compression stresses thus 
obtained should not exceed the maximum allowable stresses as 
recommended on p. 463. In computing the combined stresses, the 
full section, including the fireproofing, may be used, in the same 
manner as in beams and slabs at the support. If the stresses 
exceed the allowable unit stresses the section should be increased. 

The final step is to determine the required amount of tensile 
steel to resist the bending moment. For this purpose, the com- 
bination of the vertical column load and the bending moment which 
produces maximum tension should be used. This combination will 
be different from that used for determining maximum compression. 
There is a difference of opinion as to the method of determining 
the required amount of reinforcement to resist the bending moment 
in the column. 

The following method, developed by the authors, is recommended 
for use: 

For the top columns, where the vertical column load is small in 
comparison with the bending moment, the amount of tensile steel 
should be determined for the bending moment alone, the effect of 
the vertical load being disregarded. The amount of steel is deter- 


mined from the beam formula, A, = i where d is the distance 


from center of tensile steel to the face of the column. If the amount 
of steel required by the column design is not sufficient, additional 
bars should be used. These do not need to extend the full length of 
the column, as the bending moment decreases rapidly. 

For columns in lower floors, where the total dead load carried by 
the column is appreciable, the following method for computing 
tensile stresses in the column is recommended by the authors: 

Assume that the structure above the floor under consideration is 
loaded with dead load only, but that the floor under consideration is 
loaded with dead and live load. Under such conditions, provide 
sufficient tensile strength in the column to give a factor of safety of 
2 in case of overloading of the floor ‘with the structure above not 
loaded). 

For this purpose, compute the dead load on the column above the 
floor under consideration. Add to it double the reaction due to live 
and dead load from the floor. 


A62 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Compute the bending moment in the column produced by the 
dead and live load on the floor under consideration. | 

Multiply this moment by 2 and combine it with the column load 
computed above. Provide such an amount of steel in the column 
that the tensile stresses will not exceed double the allowable tensile 
stresses in steel. 

This method gives the same factor of safety against cracks in 
column as is used in the floor construction. 

This method will be understood from the following explanation: 
The column load above the floor under investigation is not affected 
by the loading condition in that floor. It may easily happen during 
the life of a structure, and even during construction of the structure, 
that the wall panels of one floor are overloaded, while no appreciable 
live load is placed on the floors above. With such loading, only 
the dead load above the floor can be counted upon as reducing the 
tensile stresses. This dead load is constant. It will not increase 
with the overloading of the slab. When double the design load is 
placed on the slab, the column load will be increased by the reaction 
of the slab, but the column load above the slab will not be changed. 
For this reason the constant dead load plus the reaction of the over- 
loaded slab is combined with double bending moment. 

Anchoring of Bending Steel in Wall Column.—The de in the 
column used to resist bending stresses should be properly anchored, 
else the bars may pull out instead of resisting tension. 

In the top of the top story columns, the outside bars should 
extend above the points of maximum bending moment a sufficient 
length to develop the strength of bar in tension. If this is not pos- 
sible by straight imbedment, the bars should be properly hooked at 
the top in the slab. The inside bars are compression reinforcement. 

In intermediate stories, the outside bars at the top of any column 
are sufficiently developed when they are extended into the column 
above, the proper distance from points of maximum moment to 
develop the strength of the bar in tension. The inside bars of any 
column serve as compression reinforcement below the slab and as 
tensile reinforcement above the slab. The length of imbedment of 
these bars in the column above, measured from top of slab, should be 
such as to develop the full tensile strength in the bar. 

In all above cases, an imbedment equal to 40 diameters of the bar 
for deformed bars, and 50 diameters for plain bars, may be con- 
sidered as ample. 


COLUMNS SUBJECTED TO BENDING 463 


Allowable Unit Stresses for Members Subjected to Axial Com- 
pression and Bending.—Considerable difference of opinion exists 
as to the allowable compressive unit stresses in members subjected 
to axial compression and bending. The maximum stresses are 
neither pure compression stresses nor pure bending stresses, but a 
combination of the two types. (For explanation of difference 
between the two types of stresses, see p. 30.) The proportion of 
the bending stress to the compression stress entering into the make-up 
of the maximum stress varies. It is 50 per cent or less of the total 
stress for condition of loading where no tension is developed in the sec- 
tion, and more than 50 per cent in columns where considerable tension 
is developed. The allowable compressive unit stress in the two cases. 
should be different and should be fixed according to the proportion 
of the two types of stresses. The authors’ recommendation is based 
upon this principle. 

Authors’ Recommendations.—For members subjected to axial 
compression and bending, the maximum compressive unit stress 
should not exceed values given below. 

(a) When the whole section is under compression or when only 
negligible tension is developed, the maximum compressive unit 
stress should not exceed 1.4 times the allowable unit stress for axial 
compression. With allowable compression stress f, = 0.225 f’., the 
combined stress will amount to 0.315 fc. 

(b) When considerable tension is developed in the section, the 
maximum compressive unit stress should not exceed the allowable 
unit stress in bending, namely 0.4’. 

Joint Committee Specifications, 1924.—The recommendations of 
the Joint Committee, 1924, are as quoted below: 


“Reinforced concrete columns subject to bending stresses shall 
be treated as follows: , 

‘“‘(q) With spiral reinforcement: The compressive unit stress on 
the concrete within the core area under combined axial load and 
bending shall not exceed by more than 20 per cent the value given 
for axial load by Formula (43). 

“(b) With lateral ties: Additional longitudinal reinforcement may 
be used if required, and the compressive unit stress on the concrete 
under combined axial load and bending may be increased to 0.30f’.. 
The total amount of reinforcement considered in the computations 
shall be not more than 4 per cent of the total area of the column. 

“Tension in the longitudinal reinforcement due to bending of the 
column shall not exceed 16 000 Ib. per sq. in.” 


464. CONCRETE AND REINFORCED CONCRETE COLUMNS 


Cleveland, New York, and Philadelphia Codes.—Codes of the Cities 
of Cleveland, New York, and Philadelphia specify that the stresses 
due to direct stress and bending shall not exceed in the extreme fiber 
the allowable fiber stress in bending. 

Columns Carrying Crane Loads or Loads on Brackets.—Columns 
carrying crane loads or other loads on brackets are subjected to 
bending moments produced by the eccentric position of the load in. 
respect to the center line of the column. 


Let _ P = load on the bracket, Ib. ; 
a = distance of point of application of load 
to center of column, in.; 
b = depth of bracket measured from center 
of tension to center of compression; 


1, 11, l2, = column dimensions, in. ; 
M = bending moment produced by load P, 
in.-lb.; 


Mi, Me, M3, M4 = bending moments in column at points 
indicated in corresponding figure; 


R = horizontal reaction, lb. 


Then, 
Bending Moment in the bracket Due to Load, P, 
M = Pa. sito. eee 


For cranes, the load, P, should be computed when the crane is 
in the most unfavorable position as far as the column in question is 
concerned. Sufficient allowance should be made for impact. 

The bending moments produced in the column by the bending 
moment, M, depend upon the end condition in the column, and 
also upon the position of the bracket on the column. 


The following conditions are possible: 
(1) Column fixed at the bottom and free at the top. The 
column is subjected to a bending moment equal to 
M = Pa for its whole length below the brackets. (See 
Fig. 155 (1).) 
(2) Column hinged at both ends. (See Fig. 155 (2).) 
Horizontal reaction, top and bottom, 


RPS 


COLUMNS SUBJECTED TO BENDING 465 


Moment above bracket, 


M1 = Pils eee es ime Se (BL) 


Moment below bracket, 








a 
Mz = — Poh FRO ea Pei Ba, Te 2) 
Hinged 
3 1 lg R& 
QING rie by Ve 
Pa ay 
nek 
oe ¥— ~Bracket 
Hao 
1 i} 
4 i} 
mos 
yt 
v 1 
Hie 
ith ER ded 
Fixed Hinged 
(1) Column Fixed at Bottom (2) Column Hinged at Top and Bottom 
Fixed 
Fixed Fixed 
( 3) Column Fixed at Bottom and Hinged at Top (4) Column Fixed at Top and Bottom 


Fic. 155.—Columns Carrying Loads on Brackets. (See p. 464.) 


(3) Column hinged at the top and fixed at the bottom: (See 
Fig. 155 (8)). 


Horizontal reaction, top and bottom, 


Rm aPaGl?— hb?) —bBle+ BD] . 24. « (63) 


466 CONCRETE AND REINFORCED CONCRETE COLUMNS 


Moments, 
M,=0;..f we ee eS 
Mo = 1 Ble ft a (55) 
Ms = — Pa+ Ro + 0);. +. (56) 
M, = — Pa RL 0 (57) 
(4) Column fixed at top and bottom: (See Fig. 155 (4).) 
Horizontal reaction, top and bottom, 
R= BP eal Ile +b) +h? + (e+ d)71;. . - (68) 
Moments, 
My = — Pal? — Qh + db) +2 +5)"; (68) 
Mo = M,+ Rh... ee (60) 
M3; = M, + R(b +b) — Pa; WAM es GO (61) 


M, = PSP - Gh+HQ—-h)y+ a+ . (62) 
also 
M, 


I 


M, + Ri- Pa 2 My 2 Ri 


Simplified Formulas.—The above formulas may be simplified 
without appreciable error by assuming the depth of bracket, b, equal 
zero. The formulas change to: 


(1a) Column fixed at the bottom, free at the top. 
Formulas same as for case (1). 
(2a) Column hinged at both ends. (See Fig. 156 (2a).) 
Horizontal reaction, top and bottom, 
a 


eae po (64) 
Moment above bracket, 
M, = ki: 
Moment below bracket, 
Ms. = — Rh. 


(3a) Column hinged at the top and fixed at the bottom. 
(See Fig. 156 (3a).) 


Horizontal reaction, top and bottom, 


p= 1.5P7 3(1 +7) | 3 a 


COLUMNS SUBJECTED TO BENDING 467 


Bending moments, 


Dee 0) beshioey srw ltl) a quueroyy (68) 
Mo = + Risploom) ols qo daneos, 21160) 
ee ah Rig. is eee 8070) 
YE ea i een rent A 


(4a) Column fixed at top and bottom. (See Fig. 156 (4a).) 
Horizontal reaction, top and bottom, 
a Lil 


R = 6P7 Bai ott Wire Mateus «i.e mist ate heteer « (72) 


Bending moments, 


Ms =— Pofi—4(2)4a(2)'].. . . @@ 


Mz=M,+ Rly; . ee eed Ete to 
t= NS - Rlg — Pas 
M,z=M,+ RI-— Pa = M34 Rly. 
Hinged Hinged 

R 


I 


} 





| ~ 
pero P 
1 J Mey 
eae 7 
ti M, 
as 
, 
| 
‘aig 
H | 
pe vey , 
Hinged Fixed 
(2a) (3a) (4a) 


Itc. 156.—Columns Carrying Load on Brackets, b = 0. (See p. 466.) 


The case of a rigid frame with crane brackets is treated in Vol- 
ume ITI. . 

The first condition occurs: (a) when a post is firmly anchored 
to a heavy foundation; (b) when a column extends above the floor 
and is held sufficiently at the floor level, but has no support at the 
top capable of resisting a horizontal reaction; (c) when a post is 
imbedded in firm ground well tamped but is free at the top. In the 
last case, the stability of the post depends upon the resistance of the 
soil to compression. The post compresses the earth on the side 


A468 CONCRETE AND REINFORCED CONCRETE COLUMNS 


of the bracket on the top and on the opposite side at the bottom. 
The pressure on the earth exerted by the post can be found accu- 
rately enough from the formula 


6M 


iS - ‘dh?’ U (77) 


where 
d = side of post; | 
h = depth of imbedment in ground. 


The second condition exists: (a) When the column is actually 
hinged on the top and bottom. Such cases practically never occur 
in concrete construction. (b) When the column rests on a small 
footing and carries a roof construction above, which is anchored to it 
sufficiently to resist any horizontal reaction, but the connection is 
not rigid enough to withstand bending. In such case the column is 
practically free to turn on both ends. 

The third condition takes place when a column is rigidly held 
by the floor system below and carries a roof construction anchored 
to it sufficiently to resist horizontal pull, but not rigidly connected. 
The same condition takes place when, with top free to turn, the col- 
umn is rigidly attached to a heavy foundation, or is imbedded in 
firm ground for such a distance that the resistance of the earth to com- 
pression holds the column firm. The pressure on the earth, due to 
the bending moment, can be computed as explained in connection 
with the first condition. 

The fourth condition takes place when the bottom is fixed as 
explained in connection with the third condition, and the top also is 
rigidly connected with a floor construction above. 


CHAPTER VIII 
FOUNDATIONS AND FOOTINGS 


Concrete excels as a material for foundations. Since the design 
of a foundation is governed not only by the weight of the super- 
structure, but also by the character of the supporting work or soil, 
this chapter includes a brief discussion of the method of determining 
the carrying capacity of soil and of the allowable soil pressures. 

The method of designing concrete is given in detail. 

The treatment of piles is discussed in Chapter IX. 


DETERMINATION OF CARRYING CAPACITY OF SOIL 


It is of prime importance for the safety of the structure that the 
foundation be designed with proper consideration for the conditions 
of the supporting soil. The character of the soil, therefore, must be 
carefully investigated. The most common methods are: wash 
borings, diamond drill borings, and test pits. Wash borings are 
made by driving a small pipe, up to 4 in. in diameter, and working 
a water jet inside of the pipe. The record of the material brought 
up to the surface gives a fair idea of the composition of the ground. 
The results, however, are not always sufficiently reliable, as the 
material is affected by the water. For more accurate investigation, 
Diamond drill borings are used. These are made by a sharp, hollow 
cutter which cuts through the soil and brings out a core of the mate- 
rial through which it passes. This gives reliable information on the 
conditions of the soil and the depth of the various layers. Borings, 
to be of any value, must be made by experienced operators. Test pits, 
as the term implies, consist of pits dug in the ground. ‘This method 
of exploration is too expensive where the character of the soil varies 
appreciably from point to point, as it requires a large number of test 
pits. It is restricted to shallow excavations. Test pits are often 
used in combination with borings, for more thorough investigation. 
After borings are made, the most important places are selected on 
the basis of the borings, for test pits. 

469 


470 FOUNDATIONS AND FOOTINGS 


Results of the borings should be shown on the foundation draw- 
ings, to serve as a guide for determining the size of the foundation. 
Many building codes require such a record as a part of the foundation 
plans. In addition to the borings, investigation should be made of 
the type of foundation used in the buildings in the immediate vicinity. 
In some cases, the information gained from the borings and from 
prior experience will limit the choice to one or two types of founda- 
tion. Often there is considerable latitude in the choice of foundation. 
This is the case when the layers of the materials composing the ground 
increase in hardness with the depth. 

For instance, if a layer of sand is underlaid by clay, which in turn 
rests on hardpan and rock, each of these layers could carry the foun- 
detion with an allowable unit pressure suitable for the particular 
soil. The question of depth is determined by relative economy. 
By going deeper, to a soil with large allowable soil pressure, the size 
and the cost of the footing would be decreased, but the cost of excava- 
tion would be increased. When there is danger of encountering 
water, it is cheaper to use larger footings on a higher layer than to 
go down to more solid ground. In many cases, however, the size 
and importance of the structure aré such that a foundation in unyield- 
ing soil is required and the footings are carried to hardpan or rock 
even when support could have been obtained higher at smaller cost. 
This is done because in compressible soils some settlement is unavoid- 
able, and even with carefully designed and properly proportioned - 
foundation it may be impossible to prevent uneven settlement. 

Sometimes it is found that a harder stratum of soil lies above 
much softer ground. In such a case, the foundation, if built on the 
upper layer, must be proportioned to the capacity of the underlying 
softer material. 


BEARING POWER OF SOILS 


As a general rule, a foundation should not be built on soil con- 
taining organic matter, such as loam, nor on filled or made ground. 
It should be built on firm undisturbed soil and should be of such 
dimensions that the allowable bearing on the soil is not exceeded. 

The sustaining power of soil depends upon its composition, the 
amount of water which it contains or is likely to receive, and the 
degree to which it is confined. Since the conditions vary in different 
localities, the allowable pressure on soils of the same name, as estab- 


BEARING POWER OF SOILS ATI 


lished by experience, varies to some extent. It is always advisable 
to be guided by the experience obtained in laying other foundations 
in the vicinity where the building is to be erected. 

The maximum allowable bearing values on soil, as specified by 
the Building Code of Boston, may be used as a guide in design- 
ing the foundation. These are selected because the nomenclature 
used in describing the soil is well defined, and the values have been 
arrived at as a result of broad engineering experience and numerous 
tests.! 


LTTE SS Ee gly Sa Re a 2 Ae een ee a 100 tons per sq. ft. 
UE EGR BE hs On ek ae re 10 tons per sq. ft. 
Gravel, compact sand, and hard yellow clay... 6 tons per sq. ft. 


Dry or wet sand of coarse or medium-sized 
grains, hard blue clay mixed or unmixed with 


sand, disintegrated ledge rock............. 5 tons per sq. ft. 
Medium stiff or plastic clay mixed or unmixed 

with sand, or fine-grained dry sand......... 4 tons per sq. ft. 
Fine-grained wet sand (confined)............. 3 tons per sq. ft. 


Soft clay protected against lateral displacement. 2 tons per sq. ft. 


Definitions —(a) Solid ledge: Naturally formed rock, such as the 
granites and others of similar hardness and soundness, normally 
requiring blasting for their removal. 

(b) Shale: Laminated slate or clay rocks, removable with more 
or less difficulty by picking. ) 

(c) Hardpan: A thoroughly cemented mixture of sand and peb- 
bles, or of sand, pebbles and clay, with or without a mixture of 
boulders, and difficult to remove by picking. 

(d) Gravel: A natural uncemented mixture of coarse or medium 
grained sand with a substantial amount of pebbles measuring one- 
fourth of an inch or more in diameter. 

(e) Sand (compact): Requiring picking for its removal. 

(f) Sand (loose): Requiring shoveling only. 

(g) Sand (medium grain): Individual grains readily distinguish- 
able by eye though not of pronounced size. 

(h) Sand (fine-grained): Individual grains distinguished by eye 
only with difficulty. 

(i) Hard clay: Requiring picking for its removal. 

1Of particular interest are the investigations by Mr. Joseph R. Worcester, 


the results of which are published in Journal, Boston Society of Civil Engineers, 
January, 1914, p. 19. 


472 FOUNDATIONS AND FOOTINGS 


(j) Disintegrated ledge rock: Residual deposits of decomposed 
ledge. . 

(k) Medium clay: Stiff and plastic but capable of being spaded. 

(1) Soft clay: Putty-like in consistency and changing shape 
readily under relatively slight pressure. 

If part of a structure rests upon unyielding soil, such as ledge rock 
or hardpan, and the rest on yielding soils, the allowable unit bearing 
value for the yielding soils should be reduced by from 50 to 30 per 
cent, the higher percentage to be used for soils with 2-ton capacity 
and the lower for soils with 6-ton capacity. The reason in obvious. 
Rock and hardpan are practically unyielding, and no settlement of 
the footings resting on them would take place. The footings resting 
on softer soils normally are expected to have some settlement. The 
structure resting partly on one and partly on the other material, 
therefore, would be subjected to unequal settlement. To avoid this, 
the normal settlement is reduced by reducing the allowable unit 
pressure for the yielding soil. 


GENERAL RULES OF DESIGN 


Excessive and Unequal Settlement Must be Guarded Against.— 
In designing foundations, the following facts must be borne in mind: 

(1) In yielding soils (everything softer than hardpan) some set- 
tlement of foundation is almost unavoidable. 

(2) The possibility of settlement should be reduced as much as 
possible by the use of conservative unit pressures. . 

(3) The foundation should be designed so that the settlement, if 
any, will be uniform, i.e., that all footings will settle approximately 
the same amount. 

The last requirement is of particular importance in reinforced 
concrete structures on account of the rigid connection between 
various parts of the structure: Uniform settlement, even of appre- 
ciable amount, does not produce any stresses in the members. com- 
posing the structure. When there is unequal settlement, however, 
i.e., when part of the structure settles more than the rest, even if the 
settlement is of small amount, stresses of a character and magnitude 
not anticipated by the design are produced in the various members 
of the frame. In extreme cases, the settlement of one portion may 
result in the failure of the structure. 

The first requirement will be satisfied by selection of the proper 


GENERAL RULES OF DESIGN 473 


allowable unit pressure on the soil and of an area of footing large 
enough to prevent the load from producing a pressure exceeding this. 
To satisfy the requirement for uniform settlement, it is necessary to 
proportion the area of each footing strictly according to the load the 
column resting on it is expected to carry. Often, for the sake of 
sumplicity or appearance, columns carrying different loads are made 
of the same size, the size being determined for the largest column 
load. In designing footings, this is not permissible unless the dif- 
ference in loading is inappreciable. The use of the same area of 
footing for columns carrying appreciably different loads may mean un- 
equal settlement. This, of course, does not apply where the footings 
are carried to rock or hardpan and where no settlement is expected. 

As a general rule, a new structure should not rest anywhere on 
an old foundation unless both the old and the new structures rest on 
unyielding soil. When future extension of a structure is expected, 
the outside footings are sometimes made large enough to support 
the future column load. In non-compressible soils, this would be 
permissible. In yielding soils, it often causes unequal settlement 
and more or less harmful effect, since the footings with provision for 
future loads are larger in proportion than the adjoining footings, 
which are designed only for the load they are now expected to carry. 
The end row of columns, where provision is made for future exten- 
sion, would settle less than the rest of the structure; and cracks, of a 
size depending upon the amount of settlement, would be produced in 
the end panels. After the expected addition is built, another case of 
unequal settlement may occur, as the new footings may settle more 
than the old footings, which by that time may have settled to a 
standstill. 

Proportioning Area of Footing.—The area of the footing is 
obtained by dividing the load by the allowable unit pressure on the 
soil. There are several methods of accomplishing this. 

Footings Proportioned for Dead Plus Full Live Load—The most 
common method of determining the area of the footing is by dividing 
the full column load, plus the weight of the footing, by the allowable 
unit pressure. No distinction is made in such cases, between the 
effect of the dead load and that of the live load on the settlement of 
the foundation. 

Footings Proportioned for Dead Load Only.—In this method, a 
distinction is made between the effect of the dead load and that of 
the live load. It has been observed, in compressible soils, that the 


ATA FOUNDATIONS AND FOOTINGS 


rate of settlement is large at first and then decreases as the soil 
becomes compressed. In many cases, a large part of the settlement 
takes place while the structure Is in process of construction and when 
it carries only the dead load. The effect of the dead load is thus 
larger than the effect of the total dead and live load, because the 
live load does not come until some time after the completion of the 
building. In many buildings, full loading with live load is never 
attained. In ordinary structures, the ratio of the dead load to the 
total load is much larger for outside columns than for interior col- 
umns. If the area of footings is based on the total live and dead load, 
and most of the settlement is produced by the dead load only, it 
follows that the exterior columns will settle more than the interior 
columns. To prevent for this, some engineers and also some building 
codes (notably that of the City of New York) require that, if for 
different columns in abuilding the ratio of live load todead load varies, 
the areas of footings be computed on the basis of the dead load only, 
by dividing the dead load by a reduced soil pressure obtained in the 
manner given below. The loads on the foundation for the various 
columns are tabulated, and the dead and live loads are kept sep- 
arate. The column that has the largest ratio of live load to dead 
load is then selected. For this column, the area of the footing is 
computed by dividing the total dead and live load by the maximum 
allowable soil pressure. The reduced pressure to be used in pro- 
portioning the areas of the remaining footings is then found by 
dividing the dead load for this column by the area just determined. 
The areas of other footings are finally obtained by dividing the 
dead loads carried by the footing by this reduced pressure. The 
following example illustrates this method: | 

It is desired to determine the size of footings for Columns 1 to 3, 
carrying loads given in table ‘below, for a maximum allowable soil 
pressure of 6 000 Ib. per sq. ft. | 


ee 








Dead Load Ratio 
Calin (Including Weight Live Load, Total rissa (eat 
of Footing), Ib. Ib. 
ib Load 
1 200 000 400 000 600 C00 2.00 
Z 150 000 200 000 350 000 1.33 


5) 220 000 200 000 420 000 0.91 


ne as cece 


——— 


GENERAL RULES OF DESIGN 475 


In the table above, Column 1 has the largest ratio of live load to 
dead load; hence Column 1 is used for the determination of the re- 
duced soil pressure. The required area for this column, for a maxi- 
mum allowable soil pressure of 6 000 lb. per sq. ft., is 600 900 divided 
by 6 000, equals 100 sq. ft. The dead load of 200 000 lb., divided 
by the above area of 100 sq. ft. gives a unit pressure of 2 000 Ib. 
per sq. ft., which is used as the reduced pressure on the foundation. 
The areas for the remaining footings are found by dividing their 
dead loads by reduced this pressure of 2 000 lb. 


Footing Areas 





Solin B Pa Dead Load Divided’ by| Total Load Divided by 
j Reduced Pressure, | Maximum Allowable 
sq. ft. Pressure 
1 200 000 100 100 
2 150 000 75 58 
3 220 000 110 70 





In the table just above, the dead loads of Columns 1 to 3 and the 
areas of footings computed on the basis of dead load only, using the 
reduced pressure, are given. The areas of footings that would have 
been obtained if computed by dividing the total load by the maximum 
allowable unit pressure are also given, for comparison. 

After the areas are obtained, the stresses in the footings should 
be determined on the basis of the total dead and live load, but exclu- 
sive of the weight of the footing. 

Footings Proportioned for Dead Load Plus Fraction of Live Load.— 
Some engineers vary the above method by proportioning the area of 
the footings, not for the dead load only, but for the dead load plus a 
fraction of the live load (usually one-third to one-half). The pro- 
cedure is similar to that explained above. The area of the footing is 
found for the column with the largest ratio of live load to dead load, 
by dividing the total load by the maximum allowable pressure. 
Then the reduced unit pressure is computed by dividing the dead 
load, plus the selected fraction of live load, by the area. In deter- 
mining the areas of the remaining footings, the dead load, plus the 
same fraction of the live load, is divided by the reduced unit pressure, 
This method is recommended by the authors. 


476 FOUNDATIONS AND FOOTINGS 


Footings Must be Concentric with the Column.—To get uniform 
distribution of column load on the soil, the footing must be built 
so that its center of gravity will coincide with the theoretical point 
of application of the downward load. For independent footings, 
under concentrically loaded columns, this means that the area of the 
footing must be symmetrical and its center of gravity must coin- 
cide with the center of gravity of the column. With a pile founda- 
tion, the center of gravity of the cluster of piles must coincide with the 
center of the column. If these conditions are not fulfilled, the pres- 
sure on the foundation will not be uniform, and overloading of 
certain portions, with consequent unequal settlement, will be 
induced. 

Eccentric Footings.—Footings for walls and columns on the 
property line are sometimes made with a projection on the inside 
only. The center of gravity of such footings does not coincide with 
the center of the column. ‘The pressure is eccentric, and the dis- 
tribution may be found as explained on p. 170. The maximum unit 
pressure is much larger than if the pressure were uniformly distributed. 
For an eccentricity equal to one-sixth of the width of the footing, 
the maximum pressure at outside of wall is double the average pres- 
sure and is zero at the other edge, Such design, obviously, is not 
permissible for columns carrying loads of any magnitude. It may 
be used only for walls carrying comparatively small loads, when they 
are kept from leaning outward by the pressure of the earth behind 
the wall. The active earth pressure, together with the passive 
resistance, may then be sufficient to offset the eccentricity. 

Attempts are sometimes made to offset the effect of eccentricity 
on the foundation by making the connection between the column 
and the footing rigid and designing the column strong enough to 
resist a bending moment equal to the column load multiplied by the 
distance of the center of gravity of the column from the center of the 
footing. However, no attempt is made anywhere to balance the 
bending moment supposedly developed in the column. Such pro- 
cedure is, obviously, insufficient. It is clear that it is not possible 
to develop bending in a column unless it is held rigidly somewhere, 
and unless the connection is strong enough to balance the same bend- 
ing moment. | 

For instance, the bending moment produced in the column may 
be transferred to the floor construction. The floor beam in the first 
floor would then have to be strong enough to resist this bending 


GENERAL RULES OF DESIGN 477 


moment. This is not practicable for large column loads or large 
eccentricities, as it would require excessive strength in the beams. 
The combined footing, or cantilever footing, described on p. 522, 
offers a better solution. 

It is often stated, in defense of ents footings, that the col- 
umn load gradually spreads itself, so that by the time it reaches the 
ground it acts centrally on the footing. The unsoundness of this 
argument is evident. The column load cannot change its center of 
gravity by distributing itself to the foundation. Only an additional 
force could change its position. Even if the pressure should be uni- 
formly distributed on the foundation, there would still remain the 
unbalanced bending moment produced by the load acting on the top 
of the column and the upward reaction at the bottom acting eccen- 
trically in respect to the column load. This bending moment would 
turn, or tend to turn, the structure, even if the column and footing 
were perfectly rigid. 

Proper Design of Foundation at Property Line.— When the build- 
ing line coincides with the property line, the foundation must be 
placed within the building lines. In such cases, it is often impos- 
sible to build the footings for the wall columns concentric with the 
columns. To prevent eccentric. pressure on the foundation, the 
footing for the wall column may have to be combined with the 
interior footing, see Fig. 173, p. 5380, or the eccentricity taken care 
of by a strap beam extending from the wall column footing to the 
interior footing. ‘The design of several types of such footings is 
given on pp. 533 and 538. For combined footings, it should be 
remembered that the center of gravity of the loads must coincide 
with the center of gravity of the foundations. 

Sometimes, when the footings consist of caissons or caisson piles, 
it is necessary to place them some distance within the building line. 
Then the wall column, if placed at the building line, must be carried 
on a cantilever formed by extending the strap beam between the wall 
column and the interior column. 

Anchoring.—In structures such as chimneys, narrow buildings, 
towers, trestles, and viaducts, the foundation must be designed with 
due consideration of the lateral wind pressure. If there is any pos- 
sibility of uplift, the structure must be anchored to the foundation; 
and the weight of the foundation block, plus that of the earth above, 
must be sufficient to counteract the uplift with a proper factor of 
safety. 


478 FOUNDATIONS AND FOOTINGS 


Minimum Depth of Foundations for Outside Columns.—Foun- 
dations for outside footings of a building, in localities affected by 
frost, must be carried below the frost line. In the vicinity of New 
York City, 3 to 4 ft. below ground level is considered ample. 

Footings on lines abutting adjoining property are often carried 
to a depth of, say, 9 to 10 ft. below established grade, in order that 
they may not be disturbed by possible excavations on the abutting 
property. According to law, in case of excavation on adjoining 
lot, the owner of the existing building is required to support his 
foundation if it does not extend to the depth below grade, usually 
9 to 10 ft., established by law. 

Encroachment on Public Streets.—Before designing a founda- 
tion on the line of a building adjoining the street, it is necessary to 
investigate the local regulations. Some cities grant permission to 
encroach with the foundation under the street or sidewalk, so that it 
is possible to use concentric footings for wall columns. The distance 
of permissible encroachment is usually governed by the depth of 
foundation. In Chicago, for instance, permits are issued for founda- 
tions projecting into the street, according to the depth, for as much 
as 3 ft. at an 8-ft. level below grade, with the express understanding 
however, that any foundation less than 20 ft. below city datum shall 
be removed by the owner if it interferes with projected public works, 
such as sewers, subways, etc. 

New York, on the other hand, does not allow any encroachment 
with foundations, although it allows some encroachment with such 
parts of the building the removal of which would not affect. the 
stability of the building. 

Loads Used in Design of Foundation.—The size of foundation is 
computed by dividing the load by the allowable soil pressure, as 
explained on p. 473. The load used in the determination of the area 
of the foundation or footing is the total load coming on the column 
carried by the foundation, plus the dead load of the foundation and 
the basement wall, if any. The same reduction in live load is per- 
missible for the footings as is allowed in column design. (See p. 453.) 

In computing stresses in the foundation, due to bending, the dead 
load of the footings and walls must not be included, as they produce 
no bending moments or shears in the foundation. | 

How Stresses are Developed in Concrete Footings.—Footings 
are generally designed for the upward reactions of the soil or of the 
piles supporting the footing. Since the upward reactions are not 


GENERAL RULES OF DESIGN 479 


active forces, the manner in which they produce stresses in the footing 
is not always understood. Obviously, the ground is not likely to 
rise in order to deflect the footings, and without deflection there can 
be no stresses. 

The following explanation will make the matter clear. The 
column load coming on the footing is not distributed evenly over the 
whole foundation. At first, a large portion of the load comes directly 
under the column. This tends to produce, and actually does produce, 
a small settlement of the footing directly under the column. If the 
column were not connected with the footing the settlement would 
continue until the ground compressed sufficiently to hold the load. 
Since the column is connected with the footing, the incipient down- 
ward movement of the column produces shear in the adjacent sec- 
tions and forces them down, although to a slightly less degree because 
of the bending produced. These sections in turn bring their neigh- 
boring sections into play, so that finally the whole footing presses 
on the earth and the whole footing settles. The amount of set- 
tlement is less at the edge than in the center, therefore, the footing 
assumes the shape of a curve of the same nature as the curve that 
would have been produced if the 
reactions of the ground were ac- 
tive pressures and were carried Position ~s. 

of Footing \ 
Ree A spOTt..) |. ¢ —-—------4|-—--- ---._--__. 

When a footing fails, it does 

not mean that the reactions of 








Original 





the soil move the footing upward, —geirement / A— ¥. Footing 

: j : , Cracks -’ ‘at Failure 
but that the projecting footing of Column : 
is not able to hold the column Fig. 157.—Failure of Footing. 
itself from settlement. The col- (See p. 479.) 


umn settles, while the projections 
of the footing, instead of following the column, break away from it, 
as is evident from Fig. 157, p. 479. 

Footings Must be not only Strong Enough but also Rigid.— 
From the foregoing discussion, itis evident that the footing not only 
must be strong enough to resist the upward reactions, but must be 
rigid enough to prevent the deflection under the reactions from 
becoming too great. It is obvious that if a projection of a footing 
under a loading equal to the reaction of the soil can deflect $ in., 
then, conversely, the column would have to settle 4 in. before the 
whole footing would be brought into play. With a flexible footing, 


480 FOUNDATIONS AND FOOTINGS 


the pressure would not be uniformly distributed over the foundation, 
but would be a maximum under the column and a minimum at the 
edges. The difference between maximum and minimum pressures 
increases with the flexibility of the footing. 


PLAIN CONCRETE FOOTINGS 


The simplest type of footing consists of a plain concrete slab or 
block, the base of which is determined by the allowable unit pressure 
on the soil, and the depth determined by strength of concrete. For 
walls, for instance, the footing may consist of projections on both 
sides of the wall, as shown in Fig. 158a, p. 480. For a column, an 
independent footing, of a design shown in Fig. 158, b and c, may be 





Fig. 158.—Plain Concrete Footing. (See p. 480.) 


used. If the projections of the footing from the face of the column 
are small, the design in Fig. 158b may by used. Where the depth 
and the projections of the footing are large, the cost may be decreased 
by stepping it, as shown in Fig. 158d. 

The area of the base of a footing is determined by dividing the 
superimposed load by the allowable unit pressure on the soil. The 
depth of the projections is obtained by figuring them as cantilevers 
loaded by the reaction of the soil and supported by the column, 


REINFORCED CONCRETE FOOTINGS 481 


The critical sections should be assumed at the face of the wall, column, 
or pedestal, and in stepped footings also at the face of each suc- 
cessive step. The ratio of the height of the projection to its length 
is governed by the allowable tensile stresses in plain concrete and the 
magnitude of the upward pressure. The table below gives the depth 
of the footing, h, expressed in terms of the length of the projection, l, 
for various unit pressures on soil and for an allowable tensile stress 
in concrete of 60 Ib. per sq. in., the unit stress recommended for 
use in foundations. In a stepped footing, the depths h, hi, and ho, 
corresponding to lengths of projections i, l; and lg, respectively, 
should be taken from the table. 


Required Depth h, of Plain Concrete Footing in Terms of Length of 
Projection, | 


Unit Pressure on Foundation, in Lb. per Sq. Ft. 


2 000 | 3 000 | 4.000 | 5000 | 6000 | 7000} 8000 9000 | 10000 

















h | 0.81 | 1.00} 1.21 | Peay etapa 1.61 | 1.71 1.81 1.91 


! 














} 








The table assumes that the whole footing acts as a unit. This 
is the case only when the whole depth of the footing is poured in a 
continuous operation. 


REINFORCED CONCRETE FOOTINGS 


To distribute the column load over .a large area of the ground 
without carrying the foundation in successive steps to a considerable 
depth and using a large mass of concrete, the foundation may be 
built of reinforced concrete. Reinforced concrete footings utilize 
the compressive strength of the concrete and therefore are more 
economical than the I-beam type of design formerly used.” 

Reinforced concrete footings may be divided into five groups: 
(1) wall footings; (2) independent column footings of rectangular 
or square shape; (3) combined footings carrying more than one 
column; (4) cantilever or strap footings; (5) raft foundation. 


* See Second Edition of ‘‘Concrete, Plain and Reinforced,” page 643. 


482 FOUNDATIONS AND FOOTINGS 


WaLuL FootinGs 


A wall footing, as a rule consists of a slab projecting the required 
distance on both sides of the wall as cantilevers. For small pro- 
jections, it is most economical to use plain concrete wall footing, in 
which. case the ratio of the length of the projection to the depth 
should be taken from table on p. 481. If the projections are large, 
it is cheaper to reinforce the footing. In figuring bending moments, 
each portion should be considered as a cantilever with the critical 


section at the face of the wall. The reinforcement, determined — 


from the bending moment in the usual fashion, consists of bars placed 
at the bottom of the footing at right angles to the wall. 

Special attention must be paid to bond stresses. The depth of 
the footing and the diameter of the bars must be such that the unit 
bond stress, based on the total external shear and determined by 
formulas given on p. 262, does not exceed the allowable unit stress. 
It is advantageous to use bars of small diameter. The use of 
deformed bars is advisable. 

Diagonal tension also must be considered (see p. 247). As a 
basis for figuring the diagonal tension, the shear is taken, figured at 
a distance from the wall face equal to the effective depth of the 
footing. It is preferable to design the wall footings of such dimen- 
sions as to avoid the use of diagonal tension reinforcement. 

In stepped footings, the steps must be of such depth that at no 
point of the footing the tensile stresses in steel, the bond stresses, or 
the diagonal tension in the concrete will exceed the allowable working 
unit stress. | 

Example of Reinforced Wall Footing. 


Example 1.—Design a footing for a concrete wall 10 feet high, 14 inches thick, 
carrying a concentric superimposed load of 26 500 lb. per lin. ft. The allowable 
pressure on the soil is 5 000 lb. per sq. ft. The allowable stresses are fc = 650, 
fs; = 16000, n = 15, u = 100 and v = 40. 


Solution. —The weight of wall is 10 x 44 X 150 = 1750 1b. per lin. ft. and the 
assumed weight of footing 950 lb. per lin. ft. Since the wall carries a load of 
26 500 Ib. per lin. ft., the total load on foundation is 26 500 + 1 750 + 950 = 
29 200 Ib. With an allowable unit soil pressure of 5 000 lb., the footing requires 
an area equal to 29 200 + 5 000 = 5.84 sq. ft. per lin. ft. of wall. A continuous 
footing 5 ft. 10 in. wide will be accepted. This gives a projection L =2.ft4 am 
on each side of the wall. If plain concrete were used, the required depth of the 
projection would be 3 ft. because, from table on p. 481, for unit pressure of 5 000 
Ib., h = 1.31 = 1.3 X 2.33 = 3.08 ft. 


= = =_ - 





REINFORCED CONCRETE FOOTINGS 483 


Depth of Reinforced Concrete Footing.—The total upward resisting pressure 
is 5 000 lb. per sq. ft. From this the weight of the footing may be deducted, 
leaving a net upward pressure of 4 840 lb. per sq. ft. The shear at the edge of 
the wall is 2.33 X 4 840 = 11 290 Ib. per lin. ft. The required depth for punching 

11 290 
12 X 120 
= 7.85. This depth is less than required for other stresses. In wall footings, 
punching shear does not need to be considered. 

The cantilever is 2 ft. 4 in. long and loaded with 4 840 lb. per sq. ft. The 
bending moment, therefore, is M = 2.33 X 4840 X 14 = 158 000 in.-lb. per 
lin. foot. The required depth, using the slab formula on p. 208 (d = 0.028 
4/M), d = 0.028+/158 000 = 11.13 in. This depth to steel is larger than 
required for punching shear and will be tried. 

Area of Steel.—Using f, = 16 000 lb. per sq. in., d = 11.13 in., 7 = 0.875, 


shear, for an allowable unit punching shear of 120 Ib. per sq. in., isd = 





the area of steel from formula is, A; = ——. 
fsgd 

2 158 000 

~ 16000 X 0.875 X 11.13 





8 


= 1.01 sq. in. per lin. ft. | 


3_in. round bars spaced 54 in. on center give an area of 1.01 sq. in. 

Bond Stresses.—The external shear at the edge of the wall is 11 290 lb. per 
lin. ft. With 3-in. round bars, 5} in. o.c., there are 2.28 bars per foot. The 
perimeter of a 4-in. round bar is 2.36. The perimeters of bars per foot equal 


Do = 2.28 X 2.36 = 5.38 sq. in. and the bond stress from formula u = =a 
oj 
11 290 


tea 660,875.06 11. 13 


= 215 lb.-per sq. in. 


The bond stress is too large. To reduce it, any of the following methods may be 
used: (1) smaller bars may be used; (2) area of steel may be increased and same 
depth retained; (3) depth of footing may be increased and same area of steel 
retained; (4) bars may be provided with hooks (see p. 263). Any combination 
of the methods may be used. 

In this case, the first method will be tried. Five-eighth inch round bars 3{-in. 
on centers (3.70 bars per foot) will be substituted for {-in. round bars 5¢ in. 0.¢. 
This gives an area of steel per foot of wall 0.307 X 3.70 = 1.13 sq. in. The 
perimeter of a 8-in. round bar is 1.96 in. The perimeter of bars per foot is 
Yo = 3.70 X 1.96 = 7.25 and the bond stress 


11 290 


= = 160. 
7.25 X 0.875 & IT.138 


This is still too large. As it would not be practicable to use bars smaller 
than 3-in., since their spacing would be too small, it will be necessary to increase 
the effective depth in the ratio of +89, where 160 is the computed stress and 100 
the allowable stress. The increased depth is 11.13 X +§§ = 17.81. To this 
must be added the protective cover. The total depth then.is 17.81 + 3.5 = 


484 FOUNDATIONS AND FOOTINGS 


21.31 in. Keeping the same reinforcement, with the increased depth, the unit 
bond stresses will be reduced to 100 lb. per sq. in. It is obvious that in such a 
case the full value of the steel in tension is not utilized, but is needed to satisfy 
the bond requirements. 
Diagonal Tension.— Diagonal tension will be figured at a section distant from 
the wall a length equal to the effective depth. (See Fig. 159.) This section is 
marked 1-1 in the figure. Since the upward unit pressure is 4 840 Ib. per sq. ft., 
the external shear there is V = 4840 X 0.83 = 4 020 lb. and the unit shearing 
4 020 | 
180,875 OO 
unit shearing stress and the depth is satisfactory. 
Final design. 


stress, 0 — 21.2 lb. This value is less than the allowable 


d= sin h = 214 it. 5_in. round bars 3% in. oO. ¢. 





Plan 








Peete eng 
His sag 
ve5 
Ss : = D Ft 4" <= 
eae es a 
Ae 
Ror ce oe 
oe sd 
Ss Vesa ete 
AA eS A eras RPO Tee ET 
Stree a Sout? Bias Fe hile Gang Palen ee ¢ iE es ea ear 
Bitsy Lae ay aaa aay Aer ed SOT MES ae i 
¥ ° i pr aie at on 
. * a2 ” , 
: 
? 
. ° 


e in eee 
See pi gmias me SRS ern. 
2 "Round, 5 Ft. 4” Ig. 
ee eR a he te a ees cn 
“Bowe on oan . Pea 


| 






rie 
‘ae 











27 







--~ d=18¢-- 


S 
Gj 
cea ee Ree 
( 
3 9 Ais ~. 
i 


-----h 


Bites pei eb 
RYISTIAY ISR IRA, 
“< 5000 lbs per sq. ft. 
1 
—>| 3’k----------—-+----- 5 Ft, 4'------------------- 3 
Geman penn nnn -- a= 5 Ft, 10 ------- ~~~ 
Section 


Fig. 159.—Details of Reinforced Concrete Wall Footing. (See p. 484.) 


Stepped and Sloped Wall Footing.—The diagonal tension com- 
puted above is much smaller than the allowable diagonal tension. 
The thickness of the footing may be reduced at the edges without 
exceeding the working stresses, either by stepping as in Fig. 160, or 





REINFORCED CONCRETE FOOTINGS 485 
by sloping as in Fig. 161, p. 485. The principle of stepping and 


sloping is explained thoroughly in connection with independent 
footings. 


ROSSA SEC 
ee ‘ 


37. 





Section 


Fiae. 160.—Stepped Wall Footing. (See p. 484.) 


Sometimes, instead of stepping, the footing is sloped as shown in 


Fig. 161, p. 485. The slope must be such that the stresses at the inter- 
mediate points are not excessive. 








gine SS ik Ue OR 
VLTENIIR ST, 





A 
STEWS 


d. bar 5 Ft. 4" Ig, gd on centers “= 


Section 


Fig. 161.—Sloped Wall Footing, (See p. 485.) 


INDEPENDENT CoLUMN FOooTINGS 


Independent column footings are of several types. The simplest 
type is shown in Fig. 167, p. 504. It consists of a rectangular or 
square slab of uniform thickness, reinforced in two or four directions. 
This type is economical only for small loads. 


486 FOUNDATIONS AND FOOTINGS 


For larger loads, where the required depth is large, the stepped 
footing is more economical. ‘This design is shown in Fig. 168, p. 507. 
The sum of the depths of all steps is equal to the required depth 
for the footing. This footing is designed in the same fashion as the 
slab type, except that it is necessary to investigate the tensile shearing 
and bond stresses at each step. 

Pedestals.—The standard design of footings used by the authors 
and recommended for general use is provided with a pedestal which 
reaches from the top of the footing up to the bottom of the basement 
slab. (See Fig. 162.) The pedestal is of plain concrete and its cross 
section is sufficient to carry the column load with a unit stress of 
450 lb. per sq. in. Since the pedestal is in the ground, no fireproofing 


e 





Top of Basement Floor -. 
SANS OE OM ATO ONS OV ey 


EE 
OB Not OB GSAt oar eF 
WHE Wisy 
UE) a 


Fic. 162.—Pedestals below Basement Floor. (See p. 486.) 


Sa Se Poa Wah Dare ie oh tenon 





|<-- Pedestal os : 


VTRY 
RYT AWN ANAM 


is necessary and its full cross section may be considered as effective. 
There are three reasons why the use of a pedestal is desirable. 

First, having a larger circumference than the column above, it 
decreases the depth of footing required for punching shear, also the 
bending moment. 

Second, it simplifies the construction when the top of the founda- 
tion is some distance below the basement slab, and especially when 
the tops of the footings for any reason are not on the same level. 
- In such a case, if basement columns were placed directly on the 
footings, their lengths would vary with the varying distance from the 
top of the footings to the level of the basement slab. The pedestals 
are made of such height that their tops reach to the level of the bottom 
of the basement slab, so that all the basement columns may be made 
of the same length. The column steel and column forms can then 





REINFORCED CONCRETE FOOTINGS 487 


be made of the same length and can be ordered ahead of time without 
regard to the depth to which it will be necessary to carry the footings. 

Third, when the top of the footing is some distance below the 
basement, the use of the pedestal is economical because it replaces 
the expensive highly reinforced column by cheaper plain concrete. 

If the ratio of the height of the pedestal to its width exceeds the 
limit for plain concrete, the pedestal must either be reinforced or its 
width must be increased. The minimum height of the pedestal 
must be such that the sum of the depth of the footing and the min- 
imum height of the pedestal is not smaller than required by punching 
shear at the edge of the column. (For explanation of punching, see 
p. 492.) 

Dowels.—Bars, or dowels, of the same number and size as in 
the column above, should be imbedded in the pedestal or footing. 
When there is no chance of tensile stresses due to bending in the 
column, the dowels should extend 30 diameters into the pedestal 
(or footing) and the same length above the top of the pedestal, to be 
imbedded in the column. When considerable tension stresses are 
possible, the length of dowels should be governed by the tensile 
strength of bar. It is of no importance whether the column bars are 
placed immediately next to the dowels or whether they are placed 
-a few inches apart. The stress in either case is not transferred 
directly from-the bars to the dowels but through the concrete. 

If the column above is of rich mix and spirally reinforced, the 
stress transmitted to the pedestal at the area of contact may be 
larger than the allowable bearing stress for plain concrete. (For 
bearing stresses, see p. 270.) For this reason, it may be advisable 
to extend the spiral into the pedestal a sufficient distance to grad- 
ually transmit the high pressure from the column into the body of 
the pedestal. Some designers accomplish this object by placing 
several layers of lateral hoops in the upper part of the pedestal. , 

Stepped Footings.—The bending moments and shears in a 
footing are a maximum at the edge of the column and decrease toward 
the ends. Therefore, it is often economical to reduce the thickness 
of the footing by stepping. To maintain the same factor of safety 
at all points, the reduction in the depth of the footing should not be 
larger than the reduction in the stresses. The problem is compli- 
cated by the number of stresses to be considered, not all of which 
decrease in the same ratio. These stresses are: punching shear 
(if considered), bending moment, bond, and diagonal tension. 


A488 FOUNDATIONS AND FOOTINGS 


Figure 163 (a), (b), and (c), for the three ratios 0.2, 0.3 and 0.4 
of width of pedestal (or column) to width of base, illustrates the 
rate of decrease of the various stresses and may serve as a basis 
for dimensioning the blocks in stepped footings. In this discussion, 
the footing is assumed to be provided with a pedestal. If the 
pedestal is absent, it should be considered as replaced by the 
column. ) 

In each of the three cases, the depths of footing required by punch- 
ing shear, by the bending moments, and by the bond stresses (the 
measure of which is the external shear) were computed at various 
points and plotted in the diagram. To get a direct comparison 
between the rates of decrease, the maximum values were accepted 
as a unit and the intermediate values as fractions of this unit. The 
variation of each item is shown by the curves starting at the edge of 
the pedestal from a common point. 

From the figures, it will be noted that the bending moments and 
the depths required by punching shear decrease in about the same 
ratio. The external shear, however, which is the measure of the 
bond stresses, decreases much more slowly. The depth required 
by diagonal tension is in the same yatio to the depth required by 
punching shear as the allowable shearing unit stress for plain con- | 
crete is to unit punching shear. However, diagonal tension does: 
not need to be considered nearer to the column edge than the depth 
of the footing (see p. 502). Therefore, it is seldom a determining 
item. 

Width of Blocks in Stepped Footings.—The curves showing the 
variation of stresses and bending moments may be used for the 
purpose of fixing the width of the blocks. The use of two blocks is 
recommended, as shown in Figs. 164 and 168. Also, it is assumed 
that a pedestal, as described on p. 486, isused. The thickness of the 
bottom block above the level of reinforcement is accepted as SIxX- 
tenths of the total effective depth, and the thickness of the top block 
as four-tenths of the total effective depth. 

The width of the top block depends upon whether the footing 
is designed so that the bond stresses are a maximum at the edge of 
the pedestal or at the edge of the top block. If the bond stresses 
at the edge of the pedestal are equal to the maximum. allowable 
working stresses, then the external shear curve governs the width 
of the top block and it must come outside of the shear curve, as 
indicated by the dash lines. 


REINFORCED CONCRETE FOOTINGS 489 































































BN . 





RS 


\ 
Ye 
i 1) 
<I 
Is 
cy 
Dai 






















































































































i 

ee : +f bf ReipFardcerler |_| 
ae he et cee sae 
Petr bas et tet O4a—t + 




































ESE 
os 
: 
ia 
ii 
fa 
ne 
Bel 
Pet 
: 





























7 , 
aa Fad CS) 
Ete of Reipforcamend | | | 









(<c) 
Fig. 163,—Variation of Stresses in Footings. (See p. 488.) 


490 FOUNDATIONS AND FOOTINGS 


Usually, it is possible to make the top block of the smaller width 
required by the bending moment curve, and then arrange the steel 
so that the bond stresses at the edge of the top block do not exceed 
the allowable unit stresses. In such cases, the bond stresses at the 
edge of the pedestal will be smaller than the maximum allowable 
bond stresses. This method should be used, where possible. |The 
amount of steel required at the edge of the top block will be the same 
as required at the edge of the pedestal, because the reduction in 
bending moments is the same as the reduction in depth. The width 
of the top block, determined on this basis, is given in table below. 

If more than two blocks are used, their width should be fixed in 
the same manner as given above. 





-- eT 


Stepped Sloped 


Fic. 164.—Standard Design of Footings. (See p. 490.) 


Dimensions of Independent Stepped and Sloped Square Footings 




















RaMaecidene Stepped Footing Sloped Footing 
Pedestal to 
Fide of Base, | Width of top block, Width at top, pce a - Pie 
Cc : : above stee 
i a, (Fig. 164 Fig. 164 Pak 
5 lined ee a Wig (Fig. 164) - 
0.10 0.36a 0.30a | 0.2d 
0.15 0.38a 0.3la 0.2d 
0.20 0.40a 0.32a 0.2d 
0.25 0.45a 0.34a ~OVZd 
0.30 0.50a 0.35a 0.2d 
0.35 0.52a 0.45a 0.2d 
0.40 0.55a 0.45a 0.2d 








REINFORCED CONCRETE FOOTINGS 49] 


Sloped Footings.—In sloped footings, the area of steel is deter- 
mined at the edge of the pedestal (or column, if no pedestal is used). 
As the bond stresses there are likely to be near the maximum allowable 
value, the slope must be such as not to intersect the shear curve. 
The slope recommended for use, shown in Fig. 164, p. 490, is tangent 
to the shear curve. 

Comparison of Sloped and Stepped Footings.—The sloped footing 
requires less concrete than the stepped footing. The reduction of 
stresses in the sloped footing is gradual. In spite of these two 
advantages, the use of sloped footing is not recommended, because 
the formwork required is more expensive and its increased cost more 
than balances any saving in material. Also, during concreting, the 
forms for sloped footing must be weighted down to prevent their 
being lifted by the concrete. 

With the information given in the above figures and in table 
on p. 490, it is possible to design a sloped or stepped footing just 
as easily as a footing consisting of a slab only. 


DESIGN OF INDEPENDENT FOOTING 


In designing independent footings, proceed as follows: 

Determine the area of base of footing by dividing the column 
load, plus the assumed weight of the footing, by the allowable unit 
pressure on the soil. (If the area of base is determined on the basis 
of dead load only, proceed as explained on p. 473.) 

Determine the area of the pedestal. 

Determine the depth of footing required by punching shear, using 
the column load exclusive of weight of footing. (See p. 492 for dis- 
cussion of punching shear.) 

Select the type of footing to be used, i.e., whether single slab, 
stepped or sloped. The dimensions of stepped or sloped footing 
may be taken from table on p. 490. 

Compute diagonal tension, to determine whether the assumed 
thicknesses at the various points are sufficient. 

Compute the bending moments and the required amount of steel. 
In stepped footing, this should be done at each step unless the blocks 
‘are proportioned as shown in Fig. 164. 

Compute bond stresses in steel, 


492 FOUNDATIONS AND FOOTINGS 


Explanation of Punching Shear.—The heavy concentrated load 
on a column has a tendency to punch through the footing, by pro- 
ducing in it a hole of a size somewhat larger than the section of the 
column. ‘To prevent this, the depth of the footing must be large 
enough to keep the punching shear within working limits. 

There is a difference of opinion as to the formula to be used in 
determining the punching shear. Some rules require that punching 


stresses be computed by formula v = - and others by formula 


v where V is external shear, b is perimeter of column, and d 


hans 
= ia 
is depth to reinforced steel in footing. The former method is more 
reasonable and is recommended by the authors. 

There is also a difference of opinion as to the working unit stresses. 
Most rules and building codes require that the unit shear in punch- 
ing should not exceed the maxi- 
mum shearing unit stress allowed 
for diagonal tension in beams 
with web reinforcement. This, 
for 2.000-lb. concrete, is usually 
taken at 120 lb. per sq. im. 
Recent tests, however, show that 
the strength in punching is much 
higher than at first assumed, and 
Fic. 165.—Failure Caused by Punching. that it is practically impossible 

(See p. 492.) to cause failure by punching 

shear because the footing would 

fail first from other causes. Therefore, the more modern rules 

(among them the report of the Joint Committee on Concrete and 

Reinforced Concrete, 1924) have no provision for computing punch- 

ing shear. [or square and rectangular footings, the general formula 
given above may be replaced by special formulas. 

Since many building codes require computations for punching 
shear, and since, with proper punching unit stresses, a satisfactory 
depth of footing is obtained, the computations are retained by the 
authors and the recommended general formulas for computing punch- 
ing shear are given below. 


Column--: 






rs Footing - 


Punching Unit Stress, 
Us bal PPMP: dye pe Ob es (1) 


¢ 


REINFORCED CONCRETE FOOTINGS 493 


Depth of Footing Governed by Punching Stress, 


V 
har + By Siden Stat oc) Mt Mea mst! Ue (2) 


The recommended unit stress for 2 000-lb. concrete is v = 120 lb. 
per sq. in. 

(Do not confuse b used in this formula for perimeter of column 
with 6 used in other formulas for width of rectangular beam.) 

This recommendation is made with the understanding that the 
depth obtained from Formula (2) should be used as a guide only, 
unless otherwise required by building code, and that, if more eco- 
nomical, a smaller depth may be used if proper provision is made for 
tensile stresses, diagonal tension, and bond stresses. ; 

Depth of Square and Rectangular Footings as Determined by 
Punching Shear.—For square and rectangular footings, the general 
formula given above may be replaced by special formulas. When 
the depth of the footing is given, allowable punching unit shear is 
obtained by dividing the total shear at the edge of the column or 
pedestal by the allowable unit punching shear times the perimeter 
of the column or pedestal. The external shear, V, to be used may 
be obtained by multiplying the column load, P, by the ratio of the 
area outside of the column or pedestal to the total area of the footing. 

Let P = column load, lb.; 

c = side of square pedestal (or column) or 0.78 diameter of 
round column, in.; 
c; = side of tema HS 


3 js ‘i ji Cc CC, 
a and a; = sides of base of footing (in ratios fe and aq, Use all 
1 


values in same units, preferably in feet) ; 
v = allowable punching unit stress, lb. per sq. in. 
Then for square footing * the external shear is 


v= P[r-(4\'], Sea atone pe boty 


and the perimeter b = 4c and for ieee footing 


v=P(- EN oy, eee esol tee A) 


aay, 


3 The total area of footing is a? square feet, the area of pedestal or column is 





E a? = C2 C 2 
c® square feet, area outside the pedestal a? — c?, the ratio = ag tle “) : 
a a 


Derivation for rectangular footings is similar. 


A494 FOUNDATIONS AND FOOTINGS 


and the perimeter b = 2(c + c1). Formula (2) then changes to: 
Depth Required by Punching, 


=i (2) |Pae Mmreee ie 


Square footing, 


also, if 





eee Sele area 


Square Footing Rectangular Footing 


Fic. 166.—Square and Rectangular Column Footings. (See p. 495.) 


Rectangular footings 





i, eed) 1 
d = (1 oo) Pr cae (8) 
also, if 
ipl, jee 
Cre cs 3(2 a 2), 
ee A (9) 


Values of constants C1 and Cz may be taken from tables on 
page 495. 


REINFORCED CONCRETE FOOTINGS 495 


1 2 
Values of Cy = | ~ (©) | 
4 a 


| 
i 0.1/0. 15/0. 20/0. 25]0.30)0.35)0. 40/0. 45/0. 50/0.55|0.60/0.65/0. 70/0. 75 














Cn |0.25/0.24/0. 24/0. 23/0. 23'0.22/0.21/0.20/0.19/0.17/0.16/0.14/0.13/0.11 
































| 1 
Values of Cy = (2 = Se) 


C1 
Values of — 
ay 























pee tei) 0227) 0.25) 0.3 | 0.35) 0.4 (0.45) 0.5/0.55! 0.6/0.65) 0.7 
1 | 0.49) 0.49) 0.49) 0.49) 0.48) 0.48) 0.48/0.48/0.47/0.47/0. 47/0. 47/0. 46 
15 ; 0.49} 0.49) 0.48) 0.48) 0.48) 0.47) 0.47/0.47/0. 46/0. 46/0. 45/0.45)0.45 © 
20 | 0.49) 0.48) 0.48} 0.47; 0.47) 0.46} 0.46/0.45/0. 45/0. 44/0.44)0. 43/0. 43 
25 | 0.49] 0.48] 0.47} 0.47/ 0.46} 0.46) 0.45/0. 44/0. 44/0. 43/0. 42/0. 42/0. 41 
30 | 0.48] 0.48} 0.47) 0.46) 0.45) 0.45} 0.44/0.43)0.42/0.42/0.41/0.40/0.39 
35 | 0.48] 0.47) 0.46} 0.46) 0.45) 0.44) 0.43)0.42/0.41/0.40/0.39/0.39/0.38 
40 | 0.48} 0.47) 0.46; 0.45) 0.44) 0.43) 0.42/0.41/0.40/0.39/0.38/0.37/0.36 
45 | 0.48) 0.47) 0.45] 0.44) 0.43) 0.42) 0.41/0.40/0.39)/0.38/0.36/0.35,0.34 
50 | 0.47) 0.46) 0.45] 0.44| 0.42) 0.41] 0.40/0.39/0.37/0.36/0.35/0.34/0.32 
55 | 0.47] 0.46] 0.44) 0.43) 0.42) 0.40) 0.39/0.38)/0.36/0.35/0.33/0.32/0.31 
60 | 0.47} 0.45) 0.44) 0.42) 0.41) 0.39} 0.38/0.36/0.35/0.33)0.32/0.30/0.29 
65 | 0.47) 0.45) 0.43) 0.42) 0.40) 0.39) 0.37/0.35/0.34/0.32/0.30)0.29/0.27 
70 | 0.46} 0.45) 0.43) 0.41) 0.39) 0.38) 0.36/0.34/0.32/0.31/0.29)0.27/0.25 























Bending Moments in Footings.—The footing is a circumferential 
cantilever and is subjected to radial and circumferential bending 
moments in the same manner as the portion of a flat slab at the col- 
umn. While exact bending moments cannot be determined by 
simple statics, the simplified method given below gives satisfactory 
results for practical use. Although only moments perpendicular to 
the sections are figured (i.e., radial moments), the circumferential 
moments are taken care of by distribution of the reinforcement over 
the whole width of the footing. By utilizing the constants in the 
tables which follow for assistance in computing the bending moments, 
the work of designing is made very easy. 

Bending Moments for Design.—A footing acts as a cantilever 
Supported at the column and loaded uniformly by the upward 


496 FOUNDATIONS AND FOOTINGS 


reaction of the soil. Referring to Fig. 166, p. 494, the bending 
moment caused by the upward reactions of the soil is a maximum at 
the edge of the column or pedestal. It may be determined by con- 
sidering the footing as cut along the diagonal lines running from the 
corners of the pedestal or column to the corners of the footing. 
This divides the footing into four trapezoids, one of which is marked in 
Fig. 166 by 1, 2, 3, 6. The maximum bending moments about the 
edge of the pedestal may be found by multiplying the reactions of 
the soil acting on the trapezoid by the distance between its center of 
gravity and the edge of the pedestal, or by computing the bending 
moments of the simple component parts of the trapezoid, namely, 
the rectangle 1, 2, 4, 5, and the triangles 1, 3, 4, and 2, 5, 6. The 
bending moment of the rectangle equals the load multiplied by one- 





half the height of the rectangle, which in this case is o ri ©. The 


moment arm for the load on the triangles 1s ve 7 : 





The reaction of the soil on the trapezoid, or on its component parts, 
equals the respective area multiplied by the unit reaction of the soil, 
which in turn equals the column load divided by the total area of the 
base of the footing. The dead load of the footing 1s not included in 
the column load, as it does not produce any bending moment, the 
upward reaction produced by it being counteracted by the weight of 
the footing. 


The maximum bending moment may be expressed by the following — 


formulas: 
Square Footing.— lor square footings. 


Let a = side of square base of footing; 
c = side of square pedestal or column (for round column 
c = 0.78 times diameter of column) ; 
P = total column load; 
Cy3 = constant. 


| 


Then, 
Bending Moment at Edge of Pedestal (or Column), 
s Cs c 
u=g(1-3 @+9)Po 9 


or : 
M = Ci3Pa. Re bef : : é fica SRP Tee (11) 


REINFORCED CONCRETE FOOTINGS 497 


The bending moment is in the same units as P and a. If P is in 
Ib. and a in in., M is in in.-lb. 
The constant C,3 may be taken from the table below, corre- 


sponding to the proper ratio of : In computing ratio = the values 
of a and c must be in the same units, preferably in feet. 


Constants C;; for Square Footing 


To be used in Formula (11) for bending moment, M = Cy3Pa (See p. 496.) 








Ratio of Side of Pedestal (or Column) to Side of Footing, = 





\ { | 
0.10) 0.15) 0.20 


i 
0.25) 0.30) 0.35) 0.40) 0.45 





| | | 
0.50| 0.60] 0.70| 0.80] 0.90 








Crs 10.071/0.065 ae: Beet 0.036)0.031 








0.026/0.017\0.010)0.005/0.001 





; : ae ‘ : 
For round column, c to be used in computing ratio -equals 0.78 times diameter 
a 


of column; thus, if diameter of column is 2 ft. and side of base is 3 ft., 
e078 <2 
ee 31 

must be same units. 


c 
= 0.19, and C;; = 0.60. In computing ratio -, both values 
a 


Rectangular Footing. 
Let a and a; = sides of rectangular base of footing; 
cand c, = widths of rectangular pedestal (or column); 
P = total column load; 
CU = consiant. 


I 


Then, 
Bending Moment in the Direction of the Side a (about Section 1-2, 


in Fig. 166), : 
1 Cc Cy 
M, = (1 ae <] (2 + °) Pa see fetuie 6 tats (12) 


ee) 1 Ch apiee ce Siem ae kel, as) teliag's CETL O) 
The moment is in the same unitsas Panda. If Pisin lb. anda 

is in in., M is in in.-lb. 
The constant C,4 may be taken from the table on p. 498 for the 


or 


proper - and =, The moment in the other direction may be found 
1 


: ° C C1 : 
by interchanging = and ft If a square column is used, c becomes 
1 


498 FOUNDATIONS AND FOOTINGS 


equal to c;. Fora round column, ¢ and ci equal 0.78 times diameter 


: . C C1 4 
of column. In computing ratios, z and mm all values must be in 
1 


same units, preferably in fect. 


Constants C,, for Rectangular Footings 


To be used for bending moment in direction a in formula, M = CyPa (See 
p. 497.) 


eee 


Ratio of One Side of Ratio of Other Side of Pedestal (or Column) to the 











Column to Corre- : ; C1 
sponding Side of Other Side of Footing, ze 
Footing, 
e 
a 0.10|0.15}0.20 | 0.25 | 0.30|0.35 0.40| 0.45 | 0.50 
2 ee pots A 2 ies ol oe 
0.10 (.071'0.07310.074/0.076|0.078)0 .079|0 .081|0.083/0.085 
0.15 0 .063\0 06510 .066/0.. 068|0. 069|0.071/0.072/0.074)0.075 
0.20 0.0560 05710 .059/0.060|0..061)0. 063|0..064|0. 065/0 067 
0.25 (0 .04910.05010.052/0.053/0 .054)0.055|0.056|0.057/0.059 
0.30 0.04310 .04410.045|0.046|0.047/0.048/0.049|0.050)0.051 
0.35 (.037/0.038/0 .039|0.040)0 .041)0.041|0.042/0.043)0.044 
0.40 0. 032/0 .032/0 .033/0.034|0..035)0.035|0 .036/0.037/0.088 
0.45 0.02710 .02710.028|0.028|0 . 029)0 .030/0.030/0.031)0.082 
0.50 ().022/0.022\0 .023|0. 0240 .024'0.024/0.025|0 .026/0.026 

















ee 


. : mre 
For round columns or pedestals, c and c; to be used in computing ratios - and 
a 


Cc ; 
= equals 0.78 times diameter of column. For square columns or pedestals, 
ay 


; ae were C1 : : 
c equals ci. In computing ratios ~ and —, all values must be in same units. 
a ay 


Tension Reinforcement.—After the bending moments are deter- 
mined, the required area of the effective steel is computed from 
formula 


M 


As = 0 B7Bdf. 


The bars composing this effective steel should be spaced uniformly 
in a width of the footing equivalent to the width of the column (or 
pedestal) plus twice the effective depth of the footing plus half of 
the remaining distance to the edge of the footing. This width can 


rod 


_ — 





REINFORCED CONCRETE FOOTINGS A499 


be expressed by 3(@ + ¢ + 2d). Additional steel should be placed 
outside of this effective width, at a spacing equal to twice the spacing 
of the effective steel. This arrangement has been found satisfactory 
from tests. 

Distributing of Steel The best method of arranging the rein- 
forcement is to place the bars in two directions at right angles, par- 
allel to the sides of the footing. The spacing of the bars should be 
uniform within the effective width. 

Another satisfactory method consists of radials and rings arranged 
in the same manner as the reinforcement at the column in Smulski 
Flat Slab System (see p. 367). 

Some designers place the steel in four directions, two rectangular 
and two diagonal directions. This arrangement gives four layers of 
steel. It is more complicated than the arrangements described above, 
and for this reason is not recommended. 

Design of Steel Placed in Four Directions.— When steel is placed 
in four directions, the same formulas for bending moments may be 
used as for steel placed in two directions. The effective area of steel 
at any section would consist of the area of bars in the band perpen- 
dicular to the section, plus the area of the diagonal bars, multiplied 
by the sine of the angle between the bars and the section. For bars 
placed at 45°, the effect of diagonal bands in each direction would 
be equal to the area of bars in one band multiplied by 1.4. 

Length of Bars——Make bars about 6 in. shorter than the dimen- 
sion of footing in the direction considered. 

Compressive stresses.— Usually, compressive stresses in footings 
are small. If it is required to compute them, the effective width in 
compression in footings of uniform thickness should be taken the 
same as the effective width for placing reinforcement. 

In stepped footings, in computing the compressive stresses, the 
width of the blocks should be taken as the width of the beam. 

Regular beam formulas, p. 207, may be used in figuring compres- 
sion where b is the width discussed above; or else the ratio of steel, p, 
may be computed, and the compressive unit stress may be deter- 
mined from this. The total effective area of reinforcement should be 
used in figuring the steel ratio, p. In stepped footings, the concrete 
area equals depth of footing multiplied by the width of the block 
in which the compression is being computed. 

Bond Stresses.—In designing footings, the most important and 
often the determining feature is the bond stresses. The rate of 


500 FOUNDATIONS AND FOOTINGS 


increase of bending moment in the cantilevers composing the footing 
is large and, therefore, large stresses in steel must be developed in a 
short distance. This produces large total bond stresses. After area 
of steel required for bending moment is found, the diameter of bars 
must be selected to provide the required total bond stresses without 
exceeding allowable working unit stresses (see p. 263). 


The bond stresses are computed by formula wu = ad (see p. 262), 


in which Zo is the sum of perimeters of the reinforcement effective 
at the section considered, jd is the moment arm, and V is the external. 
shear at the section considered. 

The bond stresses must be computed at the section of maximum 
bending moment for the same loads used in computing the maximum 
bending moment. [or stepped footings, the bond stresses must also 
be computed at the edges of each step. In sloped footings, bond 
stresses must be computed at one or two intermediate sections to 
determine whether the slope is satisfactory. 

If the bond stresses are found to be excessive, they may be 
reduced by using bars of smaller diameter, since smaller bars have a 
larger circumference for the same cross section than larger bars. 
If it is impossible to keep the bond stresses within working limits by 
the use of small bars, it may be necessary either to increase the depth 
of the footing, retaining the same steel as before, or to increase the 
amount of steel, keeping the original depth, whichever is the more 
economical. In both cases the tensile stresses in steel are reduced. 
It should be kept in mind that the allowable unit bond stress may be 
increased by hooking the bars at the ends (see p. 263). 

External Shear for Figuring Bond Stresses.—In determining the 
external shear to use in the formula for bond stresses, the footing may 
be considered as separated into trapezoids in the same manner as 
explained in connection with bending moments and illustrated in 
Fig. 166, p. 494. T he total upward pressure on each trapezoid 
is the shear producing the bending moment. It should, therefore, 
be used in computing the bond stresses. For square footings, the 
external shear equals one-fourth of the total upward load outside the 
section considered. It may be expressed by the formula below. 

Let c = side of pedestal (column or step), at edge of which shear 

is computed; 
side of square footing. 
column load, |b. 


a 
I 


I 





REINFORCED CONCRETE FOOTINGS 501 


Then 


External Shear for Bond in Square Footing, 


va=ait— (2) |e, a haese recon cies (14) 
en =4l1- (2) | 


the equation changes to 
V = GaP: . ° ° ° . . . ° ° (15) 


Making 


The value of Cy may be taken from the table on page 495. 

The same formula may be used for computing external shear to 
be used in formulas for diagonal tension, in which case the value c 
equals the side at which diagonal tension is computed. 

For rectangular footings, the external shear at side c is different 
from that at sided. It may be expressed by the following formula: 


Let a and a; = sides of footing; 
c and c; = sides of column, block, or shear section. 


Then 
External Shear for Bond Alongside c in Rectangular Footing 
vl phi Mf ati 
ee epee 
or 
Ve! Cah, irre aah 3 hoe signe? 601) 
where 


Ek: 1 Cc Ci 
Ca = (1+ 2)(1- 2p 
External shear along side c; may be obtained by interchanging ; 


— with a The value of the constant Cy; may be taken from the table 


on p. 502. 


In computing bond stresses, the values of c and c; are the sides 
of the column or block at the edge of which bond stresses are desired. 

In computing diagonal tension, the values c and c; are the shear 
sections at which diagonal tension is computed. 


502 FOUNDATIONS AND FOOTINGS 


Values of Cy; = (1 ++ :) (3 > =) 
a ay 


Values of 





eS 






































c teat | | 

‘ 0.1 | 0.15) 0.2 | 0.25} 0.3 | 0.35) 0.4 0.45 0.5|0.55| 0.6|0.65} 0.7 
Qed 0.25} 0.23] 0.22) 0.20) 0.19 0.18) 0 16/0. 150. 14/0. 120. 11/0.09|0.08 
0.15 | 0.26) 0.24] 0.23) 0.21) 0.20: 0.19) 0 17|0.16)0.14/0.13)0.11)0.10;0.08 
0.20 | 0.27| 0.25! 0.24) 0.22 0.21! 0.19) 0.18/0.16/0.15|0. 13/0. 12)0. 10)0.09° 
0.25 | 0.28] 0.26) 0.25} 0.23} 0.22) 0.20) 0 1910.17|0.15|0.14)0 12/0.11)0.09 
0.30 | 0.29) 0.27| 0.26} 0.24, 0.23) 0.21) 0 19,0.18)0.16)0.14)/0 130.11)0.10 
0.35 | 0.30) 0.29} 0.27) 0.25) 0.24) 0.22 0.20/0.18|0.17|0.15)0. 13,0. 120.10 
0.40 | 0.31} 0.30] 0.28) 0.26}; 0.24) 0.23 0.2110.1910.17|0. 16/0. 14/0.12)0.10 
0.45 | 0.33} 0.31) 0.29] 0.27) 0.25) 0.23 0. 22/0.20\0.18/0.16)0.14/0.13)0.11 
0.50 | 0.34] 0.32] 0.30) 0.28) 0.26) 0.24 0.22'0.2010.19'0.17|0.15)0.13/0.11 
0.55 | 0.35} 0.33} 0.31! 0.29) 0.27) 0.20, 0.23/0.21|0.19)0.17)/0 15|0.13)0.11 
0.60 | 0.36] 0.34} 0.32} 0.30) 0.28) 0.26 0.2410 .22/0.20/0.16)0.16)0.14)0.12 
0.65 | 0.37) 0.35| 0.33] 0.31) 0.29) 0.27 0 .25'0.23'0.20)0.18/0.16/0. 14/0. 12 
0.70 | 0.38] 0.36] 0.34] 0.32) 0.30| 0.27 0 .25)0.23|0.21/0.19|0.17/0.15)0. 138 
0.75 | 0.39] 0.37] 0.35) 0.33) 0.30) 0.28 0 .2610.24/0.22/0.20/0.17|0.15)0.18 

| 

















Diagonal Tension.—Tests indicate that in reinforced concrete 
footings, diagonal tension develops.at a distance from the face of the 
column equal to the effective depth of the footing. In figuring the 


V 
maximum diagonal tension, therefore, by the formula, v = bd (see 


p. 247), V should be taken as the upward pressure between the edge 
of the footing and a line concentric with the pedestal (or column) and 
located at a distance from the face of the pedestal (or column) equal 
to the effective depth of the footing. The value jd is the moment 
arm at the section considered, while 6 is the length of this concentric 
line. } 

For square footings, where c is the diameter of the column and d 
is the effective depth of the footing, the concentric line is a square 
with a side equal to (c + 2d). When the column is rectangular in 
section with sides c and ci, the concentric line for diagonal tension is 
a rectangle with sides equal to (¢ + 2d Jand (c; + 2d) respectively. 

For square footings, the intensity of diagonal tension is the same 
at all four sides of the concentric line. For rectangular footings, on 
the other hand, there may be considerable difference in intensity at 


REINFORCED CONCRETE FOOTINGS 503 


the different sides of the rectangular concentric line. For this reason, 
it is advisable to compute diagonal tension for each side separately, 


using for V, in formula v = the external shear at one side of the 


Ms 
bjd’ 
concentric line, and for b the length of one side of the concentric line. 
The external shear, V, at each side may be computed by means 

of Formulas (14) and (16) on p. 501, in which case for = or : and a 
1 

the ratio of the sides of the concentric line to the sides of the base of 


footing should be substituted. The tables on pp. 495 and 502 may 
be used for finding the constants C,; and C75. 
The formulas for diagonal tension are: 


. For square footings, 
Car 


ee (e+ 2djd’ Pe sea Let Bes) Be, bow Ps (18) 
where C); is from the table on p. 495, corresponding to a value in the 
table vanes (¢ + 2d) 

a a 
For rectangular footings at line parallel to column side, c, 
ements 5F 
” ~ (e+ 2d)jd’ ae 


where C;5 is taken from the table on p. 502 where the value . is equal 
c- 2d ci + 2d 
ay i 





to 
a 


and <! equal 

ay 

In the above formulas, if P is in pounds, ¢ and d in inches, the 
unit stress, v, is In pounds per square inch. 

Examples.—The following examples illustrate the method of 
designing reinforced concrete column footings. 


SIMPLE SLAB FOOTING 


Example 2.—Find the dimensions of a footing for a column 28 in. square, 
carrying 350 000 lb., when the allowable pressure on the soil is 2 tons per sq. ft. 


Solution.—Necessary area of base of footing is found by dividing the total 
superimposed load, plus assumed weight of footing (say 40 000 Ib.), by allowable 


000 
Oa us 97.5 sq. ft. as the required area of 





390 
unit pressure on soil, which gives 10 


504 FOUNDATIONS AND FOOTINGS 





base. A base 10 feet square, therefore, will be selected. The final dimensions 
and reinforcement selected, from the computations below, are shown in Fig. 167, 
p. 504. 

Find Thickness of Footing as Determined by Punching Shear.—Since the 
eross area of base of footing is 100 sq. ft., and the area of a 28-in. square column is 


















\ 
Be | 
Ee 1 
H+ Wes 
Be H 
! i 
| i] 
I \ 
i] ! 
| SED 
\ als 
=) SiN 
> Si) + 
=H eS) = 
per ite  eoe, 
ee 
Nee any 
UP Alay | 


ees | debe eee rally oe A GY, 


WY YE) /2 TAA SIRES AN ANY) we ee 
“WZ 4000 Ib. per sq. ft.'7 "7% “a || ts 
ps 7" bis " ise) 
a} \ Alot eees Equal Spaces ---7'- 11" ---~= vs -4f0 


Fic. 167.—Details of Square Footing. (See p. 504.) 


5.43 sq. ft., the area of footing outside the column is 100 — 5.43 = 94.57 sq. £6; 
The upward pressure on this area produces punching shear at the edge of the 
column, which for 1 : 2 : 4 concrete should not exceed 120 Ib. per sq. in. (See 
explanation of Punching Shear, p. 492.) The load producing punching shear 
may be determined by multiplying the total load on the footing (exclusive of the 
weight of the footing, which produces no shear) by the ratio of outside of column 


REINFORCED CONCRETE FOOTINGS 905 





: ‘ 7 
to gross area of the footing. It is, therefore, x 350 000 = 331 000 lb. 


(Same results may be obtained by multiplying the load, P, by constant from 
; CoP oo 
table on p. 495, corresponding to - = Ch 0.23.) By dividing this load by 
a 


the perimeter of the column, or 112 inches, times the allowable unit shear, we get 


331 000 


OA hier: 
112 x 120 mace 


Accept 25 inches as the effective depth of the footing. 

Find Diagonal Tension.—To determine whether this computed depth is suffi- 
cient, diagonal tension will be determined at a distance from column equal to 
- effective depth of footing, as explained on p. 502. The side of the square at which 
diagonal tension is computed is c + 2d = 28 + 2 X 25 = 78 in., or 6.5 ft. The 


BigsG ; ris 
ratio — to be used in finding the constant Cy: is ener 0.65, for which Cp, = 
a 


z(1 — 0.65?) = 0.144. (Value of Cy; may also be taken from table on page 495.) 
The value of diagonal tension from Formula (18), p. 503, is 


0-144 X 350000 _ 4. 
= = - per 7 IMs 
ex eed et 





Hence, depth is satisfactory and no shear reinforcement is necessary. 
Find Bending Moment.—As explained on p. 497, the bending moment may 
be found by the use of the table on p. 497. The ratio of diameter of column to 
Find es 28 
side of footing is OF 4008 DD 
table, by interpolation, Cy; = 0.0567. The bending moment, therefore, is from 
Formula (13), p. 497, 


= 0.2338, and the corresponding constant from the 


M = 0.0567 X 350000 X 10 X 12 = 2 380 000 in.-lb. 


Find Area of Steel.—The reinforcement will be placed in two directions, 
parallel to the sides of the footing. The area of steel in each band is found by 
dividing the bending moment determined above by the moment arm, jd, times 





M 
the allowable unit stress in steel. (Forme A,=- aj i 
IQs 
2 380 000 


JY SAGE as shekaliabanggc toe OE 
7 25 x 16 000 Bivona 


Ss 


requiring twenty-three §-in. round bars. 


All these bars must be placed within the effective width, which is 28 + (2 * 25) 
+42 =8ft.3in.4 (See p. 499.) Add one bar at each side, making a total of 
twenty-five 3-in. round bars. 


4Same result may be obtained from formula 3(a + c + 2d). (See p. 499.) 


506 FOUNDATIONS AND FOOTINGS 


Find Bond Stress.—Bond stresses are determined by Formula (50), p. 262. 
The external shear outside the column is 331 000 Ib. as determined in computing 
331 000 

4 
= 82750 lb. When shear producing punching is not known, the external shear 
‘may be found from Formula (15), p. 501, using constant Cy. The number of 
effective bars per band is twenty-three 3-in. round bars, the perimeter of which is 
23 X 1.96 = 45.1 in; therefore, the unit bond stress is 


sh 82750 
See eit): 





punching shear. The external shear at one edge of column is V = 





U = 84 lb. per sq. in. 
This bond stress may be used for deformed bars, but is somewhat excessive for 
plain bars in 1 : 2 : 4 concrete (see p. 263). Hence, use deformed bars. 

The weight of the footing does not need to be considered in figuring bending 
moment, shear, diagonal tension, and bond stresses, because it is balanced by the 
upward reaction. It increases, however, the unit pressure on the soil; therefore, 
it was considered in determining the size of the base of the footing. 


SLOPED AND STEPPED FOOTINGS 


The design shown in the previous example is not economical for 
deep footings. In such cases, either a sloped or a stepped footing 
should be selected. The design of both types of footings will be 
illustrated in the following example. 


Example 3.—Design a footing for a 40-in. round column carrying a load 
P = 1000000 lb. when the allowable soil pressure is p = 6 000 Ib. per sq. ft. 

Solution.—Area of Footing—Assume weight of footing W = 50000 lb.; 
then total load P + W = 1000000 + 50 000 = 1 050 000 lb. 
P+wW_ ~ 1050000 

_ = = (175 agit. 


6 000 
Use a = 13 ft. 3 in., making area of footing a? = 175.5 sq. it. 

Size of Pedestal.—(See p. 486.) For unit compressive stresses f, = 450 Ib. 
1 000 000 
450 

The side of the square pedestal is c = 47 in. Use 48 in. square pedestal. 
Depth for Punching Shear at Edge of Column.—Area of footing is 175.5 
314 X 3.33? 





Required area of footing is 


per sq. in., the required area of pedestal to carry load P is = 2222 8qi in. 


sq. ft. Area of column = Peg 8.7 sq. ft. Area outside the column 
; . 166.8 ; 
is 175.5 —8.7 = 166.8 sq. ft. Ratio i755 = 0.95. Load to be used in com- 


puting depth is 0.95 P = 950 0001b. Since circumference of column is 40 X 3.14 
= 125.6 in. and allowable shear v = 120 lb. per sq. in., 

_ 950 000 

p9 25y69<) 120 





== Go 10g 


REINFORCED CONCRETE FOOTINGS 507 


This is the minimum depth from top of pedestal to plane of footing reinforce- 
ment. ‘The depth could also be found by using table on page 495, as is shown 
below. 


Depth for Punching Shear at Edge of Pedestal.—Since a = 13.25 ft. and 


Cc ? : 
side of pedestal, c = 4.0ft.,- = 0.3, and1 — (<) = 0.91. The load producing 
: a a 


ee o\2 
punching is P : “a | = 0.91 P = 910000 Ib. For perimeter of pedestal, . 


4c = 4 X 48 = 192 in., and v = 120 lb. per sq. in. 


910 000 


RCE eae aeons (ee 
1 192 x 120 Zs 


allowing 33 in. from the bottom of footing to center of steel, the total depth of 
footing below pedestal h = 43 in. 


Same result for d; is obtained from table on p. 495. For es 0.3, Cr = 0.28. 
a 


u ee eg OO kag) 
ence => Tae) eS 3 ; 
eters 48 X 120 ae 







Top of Basement 
F 


x 









4Ft.0 Sq. 


S Pedestal 






i ole y ay 
Top Block = i 
| = a 
I. Spgs tae CPLR og eer na ~ ap 4 
N ' 
Bottom Block ; t 
rah 


UINNUIN WIN Gp CIN INN UNG SOON III SMINEIR SIN 
44 - 5- Round bars, each way 
warn nn rn nnn eerie 13 Ft. 8" 8q----- ---------------—---? 


Fic. 168.—Stepped Footing. (See p. 507.) 


Shape of Footing.—Standard design shown in Fig. 164, p. 490, will be used. 


The dimensions a, c, d, and d; being known, the remaining values may be taken 
from table on p. 490. 


For Stepped Footings.—The bottom block is 0.6d: = 0.6 X 39.5 = 23.7 in.; 
c 
use 234 in. The top block is 39.5 — 23.5 = 16 in. For a 0.3, from table 
on p. 490, a1 = 0.5a or 6 ft. 8 in. 


For Sloped Footing.—For ae 0.3 from table, the value of az = 0.35a, or 4 ft. 
a 
8 in. and d, = 0.2d = 8 in. 


508 FOUNDATIONS AND FOOTINGS 


Diagonal Tension.—Compute diagonal tension at a distance equal to d; from 
face of pedestal. The side of the section isc + 2d, = 48 + 79 = 127 in. = 10 it. 
7 in. The external shear to be used in computing diagonal tension may be 


10.58 
obtained by the use of table on p. 495. Call c = 10.58, a = 138.25, big yee 


a 13.25 

= 0.799, for which, by exterpolation, Cr = 0.09. . 

Diagonal Tension for Stepped Footing.—The depth of a stepped footing at 

section considered is equal to the depth of the bottom block, or 23.5 in. Hence, 
unit shear from Formula (36), p. 247, 


0.09 1 000 000 
oe DA Ee eA int 
OS 197 SC D8 B Pe 


As this is smaller than the allowable shear, the design is satisfactory. 









Top of Basement 
Floor Slab ~ 





Pedestal 


ae Nis 
~ ZS < A SS S 
1) 
” % “ 
----99,58-—-- . 


Z ae ‘yT rial Footing 
(} i 


= 


i“ 





. ‘jad 
-----39,54---— 








TONG 


ANGINA RUNING NUANCE GINUNS OS 
31-7 Round bars, each way 


Fic. 169.—Sloped Footing. (See p. 507.) 
Diagonal Tension for Sloped Footing.—The depth of a sloped footing at sec- 
tion considered is 17.7 in. Hence, unit shear 


0.09 x 1 000 000 
v= 
Wi X a et 





= 46.0 lb. per sq. in. 


With an allowable unit shear of 40 lb. per sq. in., this shear is too large. The 
slope of the footing should be changed so as to get, at the section considered, a 


46.0 
depth equal to 17.7 X ere 20.35 in. The revised slope is shown in Fig. 


169, p. 508. 
Bending Moment.—Find bending moment at edge of pedestal. The side of 


¢ 4 
pedestal, c = 4 ft., side of square footing, a = 13/25 ft, Tate = = eae: = 048 


Therefore, from table on p. 497, constant in formula M= C,,Pa® is Cy = 0.047 
5 See p. 496. 


REINFORCED CONCRETE FOOTINGS 509 


and the bending moment at edge of pedestal, 
M = 0.047 X 1000000 X 13.25 = 622 800 ft.-lb. or 7 474 000 in.-lb. 
Required area of steel for d; = 39.5 in. is 


: 7 474 000 
~ 16000 X 0.875 X 39.5 





= 13.5 sq. in. 


s 


Thirty-one 3-in. round bars give an area of 13.70 sq. in. 

Compressive Stresses in Concrete.—Width of top block is 6 ft. 8 in., or 80 in.; 
13.5 
39.5 X 80 
Since, with f; = 16 000 and n = 15, for an allowable stress of 650 lb. in concrete 
p = 0.0077, and the ratio actually used is only 0.0043, it is evident that in this 

case compression is considerably below the allowable unit stresses. 

Bond Stresses.—For sloped footings, with the slope accepted in design, the 
bond stresses should be investigated only at the edge of the pedestal where the 
external shear is the largest. For stepped footings, on the other hand, it is neces- 
sary to compute bond stresses also at the edge of the top block. 

Bond Stresses at Edge of Pedestal for Stepped and Sloped Footings.—The 
external shear at the edge of pedestal, for figuring bond, is one-quarter of the load 
for punching shear 


the depth, d = 39.5 in.; A, = 13.5 sq. in.; hence p = = 0.0043. 


1 
y= 10000 oor 5001 


The depth is 39.5 in.; perimeter of thirty-one }-in. round bars is Zo = 31 X 2.35 


= 72.8 in. Hence, the bond stresses, from formula wu = Sond’ 
Oj 


= 227 500 
~ 72.8 X 0.875 X 39.5 





u = 90.5 lb. per sq. in. 


This bond stress is satisfactory for deformed bars. For plain bars, the stresses 
should be reduced as explained on p. 264. 

Bond Stresses at Edge of Top Block for Stepped Footing.—The width of the top 
block, as seen in Fig. 168, is 6 ft. 8 in., and the depth of footing at the edge is 23.5 
in. The external shear, for figuring bond stresses according to Formula (14), 


p. 501, is 
' 6.66 \? 


The same value would be obtained from formula V = CpP, where Cp = 0.18, 
from table on p. 495. The perimeter of bars is the same as computed above, or 
Do = 72.8 in. The bond stresses, computed as above, are 


187 000 
eee ee Seon in. 
“72.8 X 0.875 X 23.5 Path a 


510 FOUNDATIONS AND FOOTINGS 


This shear is too large unless the bars are hooked at the ends. If forty-four 3-in. 
round bars are substituted for thirty-one 2-in. round bars, the perimeter changes 
to to = 441.96 = 86.2 

185 000 


Uu= = 104 lb. per sq. in. 
86.2 X 0.875 X 23.5 





This stress may be considered satisfactory for deformed bars. If further redue- 
tion of bond stresses is desired, it will be necessary to increase the width of the 
block. The use of still smaller bars would not be wise, as their spacing would be 
too small. (See also p. 264.) 


RECTANGULAR FOOTINGS 


Rectangular footings are often used for wall columns and for 
rectangular interior columns. If it is permissible to go outside of the 
building line, independent footings for wall columns will be found 
most economical. The wall columns usually being rectangular, the 
footing is also made rectangular. If not governed by other require- 
ments the shape of the footing should be made such that the length 
of the projections of the footing from the column edge on all sides are 
equal, since this requires the minimum amount of materials. The 
area of the footing is obtained first by dividing the total load on the 
foundation by the allowable unit pressure on the soil. The sides of 
the rectangle of the base of the footing, which should have equal 
projection from the edge of the column on all sides, may be obtained 
by trial or from the following formula. 





a=3e-a)+VA+i—«c)2, .. . (20) 


where a is one side of the rectangle, A is the area of base, and c and cy 
are sides of the column. The values may also be taken from table on 
tao | 

Design of Rectangular Footing —Compute the area of the footing 
by dividing total column load, plus Weel of footing, by the permis- 
sible unit pressure on the soil. 

Select sides of the rectangle which will give the required area. 
For most economical design, select such dimensions for the sides 
that the projections on all sides are about equal. Use table, p. 511. 

Compute depth of punching shear, as explained for square footings. 
(See explanation of punching shear, p. 492.) 

Check the depth for diagonal tension, as for square footing. 


1 Length of Long Side, a, of Rectangular Footing 


Eacnemics 
(c— a)V A = 


c,; = short side of column. 


area of footing, c = long side of column, 


(c — ¢1)2, where a 
? 


Bh 
4 


z 
2 


From Formula a 


long side of footing, A 


Long Side of Footing, a, in Feet 


REINFORCED CONCRETE FOOTINGS 


Values of c — c, in Inches 


511 



























































7 NNORHO OH OH DOH MDNDO 
Re) CH ODDROOHHANMMNOH HH 
oO Se I ee ee es | 
rs SserOnomMoaniwmowodnnwmrewnHwo oo 
= Cre KHORDRAOHHAHNANNMAOH SH = 
faa) See oe ee oe eB es | 
us DODMADHTAHAYTONSOH OS 
Q OnNRDAMOOHANNNAOH HS 
faa) Sen I cee I ee Be oe ee | 
: DHOHPOKRNDMODOARMDOOOL 
oO MOnMDARMRODHHANNMYONM HS 
ian) : 5 ee ee ee ce ee | 
< DOOMOCNKRAKRAOMDONMSO 
va) MOM ODOARMROOHHNANMMNOASH HSH 
N See OO coe ee I ce ee OO oe | 
p Peon own on OM D OM ONS 
Re) MOnRADADROOHHAHNANMHOOHH 
AN ee ee 
2 SoOoOtFHnTOTOUMUMMDOWOMOKRHY 
+ WMOnRWDDAROOHHNANANMNHHHSH 
N So ee oe ee ee | 
Si D199 MOrnMDMATHNHOHODOMDMOO 

AQ MOR DDAROHHHANNMNHOSH SH 
N Se OD ce oe BO re I oe ee ee | 
r: HTHNDONDDAMDOHOMOEMADOM 
oO MDOnnRORDRAOR HR HANNYOO HH 
N bn BD ce Oe ce oe BO ee ee | 
x MDOHOoOMMEMOMrRAoDOWOA 
ioe) MOnRRORAOHHHAHANANMHOHS 
coeeal Se ee Fe BO oe ee oe ee | 
z NANOCODHHONDACHDOOORNM 
Pe) WDOM~MKORDRAOOHHANNDMOH 
el Ss ee Oe ee I oe ee | 
- HHOr~rnmowonrnwootononon 
+H MOON ORMDAOOHHANNDOMH 
re bs cee ee eee oe oe eB ee | 
i SODSMANDORODOMRHDO 
aq MOOKnADADROOHHHANANMMN 
5 ae Se I ce ee BO ee ce ee OO ee ee | 
~ MPRDIMINDAHROOHONRHWDO 
SlHOOrnrnwmnananoon nt nna 
Ses OS ee A ce DO es A A oe ee | 
ie ODOrnRHHKhRMDAHOMKNOCOHHN 
oe HDOKRWADDAGADOHHNNMDHMN 
bere BD en I ee OO ee oe A cee I ee ee FI | 
‘i KROMOONKRMARAOHWMOMLN 
Ro) HO OR DDAADOHHNANANDOD 
eo wesowsws we eS NS es 
re DOOMWINAMHONDHOOHANS 
rl HAMOOnRRARMRRMROOHHANANMN 
Seen I pe OO ce I es I ce BD ee BS | 
2 MImMHHOHOHHOCOMOMK AD 
i HMDOKRKRDRBABOCOCHHHAHNANMNON 
Sn A oe A ee BD cee A ee FD ce FD ce oe (OD | 

AY 
So cmooackw S ooo So 
oH RS eo ee Ss oR eee Ss 
o rset SS oS SN es 

op 


512 FOUNDATIONS AND FOOTINGS 


Compute moments in the long and short directions. Compute 
areas of steel, and select the bars so that the bond stresses do not 
exceed the allowable values. * 

Slope or step the footing as explained in connection with square 
footings. The width of the blocks may be the same as in square 
footings having a ratio of width of footing to width of column equal 
to the ratio of the rectangular footing in the direction considered. 
The blocks will also be rectangular. 


Example 4.—Design footing for column load P = 550 000 lb., column size 
c X c, = 48 X 24 in. Allowable soil pressure p = 6 000 lb. per sq. it 


Solution. —Assume dead load of footing W = 500001b. Total load P + W = 
550 000 + 50000 = 600 000 Ib. 

Required area of footing is equal to (P + W) + p = 600 000 + 6 000 = 100 
sq. ft. The sides of the column are c = 48 in., and c, = 24 in.; hence ¢ — ¢: = 
24 in. Sides of a rectangle having an area of 100 sq. ft. and having equal projec- 
tions from the column on all sides may be found from table on p. 511. The 
long side is a = 11 ft., and consequently the short side, a, = 100 + 11 = 9.1 ft. 
or 9ft. lin. It will be exact enough for practical purposes to select a footing with 
a base of 9 ft. by 11 ft., having an area of 99 sq. ft. 

Depth for Punching Shear.—The ratio of the load producing punching to the 


8 
total load equals ( —~—*) =1—— =1-—0.08 =0.92. The load producing 
aay, 99 


punching is 550 000 X 0.92 = 506 0001b. Perimeter of column bi = 2(48 + 24) 
—144in. Hence, depth for punching shear for working stress of r = 120 lb. per 


sq. in. is (irom formula d = 


1200 /” 
506 000 ‘ 
= ——— = 29 in 
120 xX 144 
Bending Moments.—Formula for moments is M =CpPa. (See p. 497 i 

nl Lita hoe eee ~ = eae 

a 

C1 
ay = 9 ft. C1 = 2 ft ae eo 0.222 

: a1 


The constants Cy for the bending moments may be taken from table on p. 498. 
In the direction of the 11-ft. side, for : = 0.364 and Fy = 0.222," by 
interpolation Cy, = 0.037. In a similar manner, the constant toe the short side 
is found, using for : = 0,222 ana for z = 0.364. Ite 0.059. 


Bending Moments: 
Long cantilever side, M = 0.037 X 550 000 X aT 
2 688 000 in.-lb. 
Short cantilever side, M = 0.059 X 550000 X 9 
3 504 000 in.-lb. 


224 000 ft.-lb. or 


I 


292 000 ft.-lb. or 


REINFORCED CONCRETE FOOTINGS 013 


Areas of Steel.—The area of steel, computed from formula A, = a is: 
A 








For long cantilever, A ENG 6.6 1 
O cantilever, = = 6. Phat 
"0.875 X 29 X 16 000 a 
3 504 000 
For short cantilever, As: = = 8.6 sq. in. 


0.875 X 29 X 16 000 


The effective steel will consist of thirty-four 3-in. round bars 10 ft. 6 in. in | 
long direction and forty-four 3-in. round bars 8 ft. 6 in. in short direction. 
This should be placed within the effective width. Also, two additional bars on 
each side should be placed outside the effective width. 

Bond Stresses.—At edge of column, the shear on each side of the column to 
be used in figuring bond stresses may be found by determining the area of the 
trapezoid and computing the upward pressure acting on same. It may be found 
more easily by using the table on p. 502. 

Thus, for the shear along the long side of column, for which, as computed 


c é 
above for short cantilever, — = 0.364 and — = 0.222, we find a constant Cy; = 
a a1 


0.27, and the shear V = 0.27 X 550000 = 148 000 lb. 


: Cc c : 
For the shear along the short side, interchange values of — and nie using for 
a a1 


in the table 0.222 and for —! 0.364. The constant from the table then is 0.19, 
a Qa, 


and the shear V = 0.19 & 550 000 = 106 000 lb. 

The shear along the long side, which is 148 000 Ib., should be used in connec- 
tion with steel running at right angles to it, i.e., steel running in short direction. 
Circumference of forty-four 3-in. round bars is 44 X 1.57 = 69 in. Hence 


from formula u = ——, 
Zojd 
148 000 


= ——_—_______ = 8 Jb. . in. 
** 69 X 0.875 X 29 eae 


This is satisfactory for deformed bars. For plain bars, the number of bars 
. would have to be increased in the proportion of the computed stress to the allow- 
able stress, or $4, or the bars provided with hooks at both ends. In this case it 
would be cheaper to use additional bars. Forty-six }-in. round bars will be 
required if plain bars are used. 

The shear along the short side of the rectangle is 106 000 Ib., and thirty-four 
2-n. round bars are used perpendicularly to that side. The suin of the perimeters 
is Zo = 34 X 1.57 = 44 in. Using formula as above, 


106 000 
uU = 
03 X 0.875 X 29 





= 79 lb. per sq. in. 


This is satisfactory for plain and deformed bars. 

Sloped Footings.—The rules for sloping footings, given on p. 490 for square 
footings, may be applied here. Since the slope is outside of the shear curve, 
bond stresses are satisfactory. 


514 FOUNDATIONS AND FOOTINGS 


Stepped Footings.—If a stepped footing is used, the dimensions of the blocks 
should be made according to table on p. 490. The sides of the step are 6 ft. 10 in. 
and 3 ft. 10 in. The sizes of blocks conform to the bending moment line, but 
not to the external shear line. The same amount of steel as’required at the edge 
of the column is satisfactory at the edge of the step for bending moment. It is, 
however, necessary to check the bond stresses, as they will be larger at the edge 
of the block than at the edge of the column. 

Bond Stresses at the Edge of Block.—The depth of the step is 17.5 in. The 


sides of the step are 6 ft. 10 in. and 3 ft. 10 in. The values of 2 and ” for the long 
a a 


1 
6.0 3.82 
side are ane 0.55 and hae 0.43. The constant Cys for these values 
(either computed by Formula (16) or taken from table) is 0.22, and the shear 
V = 0.22 x 550000 = 120000 lb. Since twenty-nine }-in. round bars act 


perpendicularly to this section, from u = =~; 
Lojd 


120 000 


u = 58 xX 0.875 X 17.5 


= 135 lb. per sq. in. 

This value for bond stresses is too large unless the bars are hooked. In the case 
under consideration, it is not feasible to reduce bond stresses by using smaller 
diameter bars, as it would require fifty-one $-in. square bars, the spacing for 
which would be too small. If hooked bars are not desired, either a larger number 


of 3-in. square bars must be used or the width of the step must be made larger, 
whichever proves to be cheaper. 


CorRNER CoLUMN FOOTINGS 


For architectural reasons, the corner columns are often made 
L-shaped. The footing must be made concentric with the center of 





pannn-=-=-Cly ae 4 


Fic. 170.—Illustration of Corner Column. (See p. 515.) 


gravity of the column. ~ The center of gravity may be found from the 
following formula. 





REINFORCED CONCRETE FOOTINGS 515 


Let a and a; = sides of the column, inches. 
c = the thickness of column, inches. 
e = perpendicular distance of the center of gravity 
from the base a inches. (See Fig. 170, p. 514.) 
Then 


ac + a;? — ¢? 
rare. ee ete er AG 
The perpendicular distance, ¢1, from the base, a1, may be found by 
‘interchanging a with a, in the formula. 
_ The above formula may also be written in the following form, 
giving the distance from base a in terms of ay: 


C= Cre } PR rae oot ere, eS eR oe Brae, (22) 





where 


a1 + 2 =| 





a Cc 
— and — 
Qa, a1 
c /a c 
sent) + 
a 
Values of Cy, = -— 
2(1 fieek, & =) 
ai ay, 
Values of s 
a an 
Qa | 


OO OO | S|, | | SY NUL 


0.2 

0.4 | 0.433 | 0.451 | 0.468 | 0.484 

0.6 | 0.385 | 0.403 | 0.419 | 0.434 | 0.450 | 0.464) 0.477) 0.489 

0.8 | 0.350 | 0.367 | 0.383 | 0.399 | 0.414 | 0.428) 0.442) 0.454] 0.466 
1.0 | 0.322 | 0.339 | 0.356 | 0.372 | 0.387 | 0.402) 0.416} 0.430) 0.443 
1.2 | 0.300 | 0.317 | 0.334 | 0.350 | 0.366 | 0.382] 0.397] 0.411] 0.425 
1.4 | 0.282 | 0.299 | 0.316 | 0.333 | 0.350 | 0.366] 0.381} 0.396] 0.411 
1.6 | 0.266 | 0.284 | 0.302 | 0.319 | 0.336 | 0.352] 0.369] 0.384! 0.400 
1.8 | 0.253 | 0.272 | 0.290 | 0.307 | 0.325 | 0.342} 0.358] 0.375] 0.390 
2.0 | 0.242 | 0.261 | 0.279 | 0.297 | 0.315 | 0.332} 0.350] 0.366] 0.383 
2.2 | 0.233 | 0.252 | 0.270 | 0.289 | 0.307 | 0.325} 0.342) 0.359] 0.377 
2.4 | 0.225 | 0.244 | 0.263 | 0.281 | 0.300 | 0.318} 0.336] 0.353} 0.371 
2.6 | 0.217 | 0.237 | 0.256 | 0.275 | 0.298 | 0.312} 0.330) 0.348] 0.366 
2.8 | 0.211 | 0.230 | 0.250 | 0.269 | 0.288 | 0.307] 0.325! 0.344] 0.362 
3.0 | 0.205 | 0.225 | 0.244 | 0.264 | 0.283 | 0.302) 0.321) 0.340] 0.358 





| 
| 


516 FOUNDATIONS AND FOOTINGS 


The design of a corner footing does not differ materially from that 
of any other independent footing. The punching shear is usually 
small, as the perimeter of the column is large in proportion to the load 
carried. The bending moments are determined in the usual way. 


INDEPENDENT FOOTINGS WITH PILES 


If piles are used to carry the column load, the size and shape of the 
footing will depend upon the spacing of the piles. If the column is 
round or square, the piles should be placed so as to make the footing 
as nearly square as possible. In many cases, it is not possible to 
space piles so as to obtain a square footing. The shape of footing 
for different numbers of piles is shown in Fig. 171, p. 517. For dif- 
ferent spacing, the shape of the footing will remain the same, but the 
dimensions of the sides will be altered proportionally. 

For concrete piles carrying large loads per pile, the spacing of the 
piles will depend upon circumstances, as discussed on p. 542. 

Design of Footing on Piles.—In the design of footings supported 
on piles, the upward reactions of the piles should be treated as con- 
centrated loads. In figuring punching shear, only the reaction of 
piles outside of the pedestal need be considered. 

Bending moments to be used in determining the steel areas are 
equal to the reactions of the piles multiplied by their distance from 
the edge of the pedestal. The piles producing bending moments in 
the direction considered are those within the trapezoid produced by 
the corner diagonals. If the lines of the trapezoid cut any piles, one- 
half of the pile so cut should be considered as acting in each of the 
two directions. | 

Placing of Reinforcement.—The reinforcement required by the 
bending moments should be distributed uniformly over the whole 
effective width of the footing, as in the case of footings resting on 
soil, because the object of the reinforcement is to make the whole 
footing act as a unit. Such an arrangement reduces the number of 
layers and also provides reinforcement for cross bending. 

Some designers treat each line of piles separately. They con- 
centrate the reiaforcement required by the bending moment pro- 
duced by each line of piles within a width across the line of piles 
somewhat larger than the diameter of the pile. Such arrangement 
is shown in Fig. 172, p. 518. 3 


REINFORCED CONCRETE FOOTINGS 517 


“| 2.6" t= > 2-6 4 


1 Pile Footing 2 Pile Footing 





4. Pile Footing § Pile Footing 


("6 
fares ey | ig 3 , 


‘ 
, 
- 


15 Pile Footing 


ait 


42-2 % 
rid ‘ a ~ 


at 





17 Pile Footing 18 Pile Footing 19 Pile Footing 


Fic. 171.—Shape of Footings for Different Arrangement of Piles. (See p. 516.) 


518 FOUNDATIONS AND FOOTINGS 


This arrangement is wrong, for two reasons. First, there is no 
provision made for cross bending. It has been established that ina 
circumferential cantilever, such as an 
independent footing, circumferential 
stresses are developed in addition to 
radial stresses. The usual arrangement, 
where bars are placed in two directions, 
provides for these stresses even if they 
are not actually computed. In Fig. 172, 
there is no steel which could perform this 
function. 

The second objection is that, usually, 
| : to take care of bond stresses, a large num- 
Fic. 172.—Incorrect Arrange- ber of bars of small diameter must be 
ment of Reinforcement above used, and if these are concentrated in a 

Piles. (See p. 516.) small width, the spacing must be made 

very small. When, on the other hand, a 

small number of bars with larger diameter is used, the bond stresses 
are excessive, thus reducing the value of steel. 





Continuous WALL COLUMN FootiNnGs 


When it is not permissible to go outside the building line with the 
footing, independent footings cannot be used for the wall columns. 
In some such cases, it is possible to use a narrow footing of a width 
equal to the thickness of the basement column and extending the full 
distance between the columns. The basement wall may be used as 
part of the foundation. , 

Bending Moment Reinforcement.—The wall, as far as the reac- 
tion of the soil is concerned, acts as a beam extending between the 
columns. If the footing extends over several panels, the interior 
panels may be treated as continuous and the end panels as semi-con- 
tinuous. The bending moments recommended are somewhat differ- 


ent from those used for continuous beams. 


Let w = reaction of soil per lin. ft. of span in Ib.; 
1, = net span in f{t.; 
| = span, center to center of column, in ft.; then 


Negative moment at center of interior span, 
wig wll; a! 12will, Fi : 
M= Tees in-lb. 2 yap (23) 





REINFORCED CONCRETE FOOTINGS 519 


Positive moment at support, 


M, = “2 Tie eee suri cin. —lOsesoteety hs tow oe (DA) 


In exterior panels, increase the bending moment in the center 
and at the interior support by 20 per cent. 

It should be noted that the loads are acting upward. The eHatatt 
of the stresses and of steel, therefore, will be opposite to the position 
of stresses and of steel in ordinary beams. 

The design of such a footing is given in the example below. 


Example 5.—Design footings for wall columns spaced 20 ft. on centers, the 
load on which is P = 440000 Ib. The width of the column is 36 in. and the 
allowable soil pressure is p = 10 000 lb. 


Solution.—Assume weight of wall and footing W = 60000 lb. Total load 
on foundation is P + W = 440000 + 60 000 = 500000 lb. The required area 
P+W i 500 000 acne 

10 000 
Since the footing extends all the way between two adjoining columns, its length 
per column is equal to the spacing of columns, in this case 20 ft. The required 
width, therefore, is 50 + 20 = 2.5 ft. The wall column will be of the same 
thickness, to make the distribution of the load uniform over the foundation. 
Design of Wall.—The wall will be placed flush with the outside face of the 
column. The footing will be stepped inside. 

In this type of footing, the wall acts as a beam loaded by the upward reaction 
of the soil and supported at the columns. The reaction per lineal foot producing 
bending moment is obtained by dividing the column load by the length of the 
footing (i.e., the spacing of columns). The weight of wall produces pressure on 
the foundation, but does not produce any shear or bending moments in the wall. 
The reaction per lineal foot is 


440 000 + 20 = 22 000 Ib. per lin. ft. 





of the footing, 


Shear and Diagonal Tension.—The thickness of the wall is governed by the 
requirement that the unit shearing stress shall not exceed the allowable unit 
stress. With columns 36 in. mice Hn net span is 17 ft. and the external shear 
at the edge of the column is V = 4,5 X 22 000 = 187000 lb. Since the effective 
depth of the wall is 11 ft., or 132 in., and the allowable unit shearing stress v = 


V 
120 lb. per sq. in., the thickness of the wall, found from formula, b = Wd’ is 
a 187 000 a 
Reio0 ix 087s 182) 





Make wall 14 in. thick. 


Diagonal Tension Reinforcement.— Diagonal tension reinforcement consists 
of stirrups, the size and spacing of which may be determined as for a beam. 
Formula (40), p. 248, may be used. Also, tables on p. 899 will be found useful. 


5b20 FOUNDATIONS AND FOOTINGS 


Projection of the Wall.—The base of the footing is wider than the 
wall; therefore, it must project outside the wall within the building. 
The projection of the base must be made strong enough to prevent 
it from breaking off. It should be designed as a cantilever and, if 
necessary, reinforced with bars placed near the bottom at right 
angles to the wall. 

If the projection is small, it should be made thick enough to 
withstand the bending stresses without reinforcement. The ratio 
between the depth and length of the projection may be taken from 
the table on p. 481. For large projections, it is more economical 
to use reinforcement. The steel must be placed near the bottom 
at right angles to the wall. It should run to the outside face of 
the wall and there be bent up and extended far enough into the 
wall to transfer to the wall the bending moment produced by the 
cantilever. 

The moment on the cantilever produces torsion in the wall. 
This is partly offset by the earth pressure against the wall. 


Bending Moment Reinforcement.—In the case under consideration, w = 
22,000 Ib., 11 = 17 ft., andl = 20ft. The bending moments are: 
Negative bending moment in the center, 


ue a 12 X.22.000 X 17 X 20 
id 16 


Positive bending moment at the support, Mi = 22 000 X 17 XK 20 = 
7 480 000 in.-lb. 


The required area of steel can be obtained from Formula (8a), p. 204, using 
effective depth of wall as d and j as). 


5 620 000 
$ X 132 * 16 000 





= 5 620 090 in.-lb. 


Top reinforcement As = = 3.0sq.in. Use 4 — 1-in. rd. 


bars. 


; 7 480 000 ; 
Bottom reincorcement at columns = = = 4:0 sq. ins aise 
z xX 182 & 16 000 





4 — 1-in. sq. bars. 
Bond Stresses.—Shear at column edge, as computed previously, is V =187 000 

lb. Perimeter of four one-inch square bars is to = 4 X 4 = 16 in. From 

formula, u = ——, 

Lojd 

Bi 187 000 

~ 16 x 0.87 x 182 





u = 100 lb. per sq. in. 
If deformed bars are used, this is satisfactory. 

For plain bars, bond stresses must be reduced to 80 Ib. per sq. in. by using an 
increased number of smaller bars of the total required area, the perimeter of which 
equals 16 X 482 = 20 in. Nine j-in. round plain bars, having an area of 3.98 


REINFORCED CONCRETE FOOTINGS o21 


sq. in. and circumference of 21.3 in., could be used. The use of deformed bars, 
however, in this case is distinctly advantageous. 

Arrangement of Steel.—Since the wallis deep in proportion to the span, it is 
not advisable to bend any of the top reinforcement. The best method is to use 
straight bars at the top. Two one-inch round bars will extend the whole length 
of the footing and will be made 20 ft. long. Two other bars will extend 12 in. 
plus or minus beyond the points of inflection on each side. The distance between 
the points of inflection is equal to 2 of the net span of 17 ft., or 10 ft. 2 in.; there-. 
fore, the bars will be made 12 ft. 3 in. 

The negative steel placed at the bottom will extend 12 in. on each side beyond 
the + of the net span. 

Compressive Stress.—Since the ratio of the areas of steel to the area of con- 
crete is less than 0.0077, compressive stresses are smaller than the allowable 
values and need not be computed. Also, since p: is less than 0.0077, the value of 
j will exceed the value j used in the calculation. This discrepancy is on the side 
of safety, as it lowers the unit bond stress and the unit shear and reduces slightly 
the stress in the steel. 

Stresses in the Wall Due to Earth Pressure.—The design of the wall for earth 
pressure is discussed on p. 640. In the present case, the wall will be considered 
as supported at the top and bottom. The earth pressure equivalent to a fluid 
pressure of a 30-lb. fluid will be assumed. The maximum pressure at the bottom 
is 9 X 30 = 270 Ib. per sq. ft. The bending moment is found by Formula (4), 
p. 640. 

M =0.77 X 270 X 9? =16 800 in.-lb. per horizontal foot of wall. If the 
wall 14 in. thick is considered as plain concrete without reinforcement, the 
_tensile stress in concrete may be computed from the formula for homogeneous 
beams, f, = = It is 
_ 6 X 16 800 


2149-9 
Fe 19 142 


Since, in addition to being supported on: top and bottom, the wall is also sup- 
ported at the columns, the actual stresses in the wall due to the earth pressure 
are much smaller than computed above, and would come below the 40-lb. tensile 
stress allowed for plain concrete. Therefore, no steel is required except some 
reinforcement for contraction due to temperature and shrinkage. This will be 
assumed as 3-In. round bars 18 in. on centers, both ways. In addition, the wall 
is reinforced with inch reinforcement. 

Projection of the Wall.—The footing is 30 in. wide and the wall 14 in. wide. 
With wall flush on one side, the projection is 16 in. The unit reaction of the soil 
(column load divided by area) is 8 800 lb. per sq. ft. Therefore, considering the 
projection as a cantilever, the bending moment at the edge of the wall is, 


M = 8800 x 18 x 8 = 93 900 in--Ib. per ft. 


The required depth, from Formula (11), p. 208, in which for f; = 16 000 Ib. per 
Sq. in., fe = 650 Ib. per sq. in., and n = 15, Ci = 0.028 is 


d = 0.028 V93 900 = 8.5 in. 


522 FOUNDATIONS AND FOOTINGS 


The required area of steel, As = 0.0077 X 8.5 X 12 = 0.79. 
Allowing 33 in. below steel, the total depth of the projection is h = 8.5 + 
3.5 =s12 in. 


COMBINED FooTINGS 


It is sometimes necessary to combine the footings of two or more 
columns. ‘This is required, for instance, when a wall column is 
placed at the building line and it is not permissible to project the 
footing of the column outside of the building line. An independent 
footing, in such a case, would be eccentrically loaded. To insure 
even distribution of the column load on the foundation, the footing 
for the wall column may be combined with the footing of the adjoin- 
ing interior column. | 

Combined footings are sometimes used for a row of columns, in 
cases where the soil is compressible and it is desirable to avoid inde- 
pendent settlement of any one column. Thus, in the new Massa- 
chusetts Institute of Technology buildings, due to the peculiarity 
of the ground, the footings of all wall columns were made continuous 
even when independent footings were feasible. 

Combined footings may also be used where the distance between 
adjoining columns is fairly small (as in corridor columns) and it is 
cheaper to build one combined footing for both columns than two 
independent footings. 

Pedestals.—The paragraph on Pedestals, p. 486, in connection 
with independent footings, applies equally to combined footings. 

General Requirements.—To insure uniform distribution of the 
column loads on the foundation, it is absolutely necessary that the 
center of gravity of the upward reaction of the soil should coincide 
with the center of gravity of the column loads. If the footing rests 
directly on the soil, then this requirement is fulfilled if the center of 
gravity of the area of the footing coincides with the center of gravity 
of the column loads. For pile foundations, the center of gravity 
of the piles (and not necessarily that of the footing) must coincide 
with the center of gravity of the column loads. 

Shape of the Base of Combined Footing.—The shape of the base 
of the footing will depend upon conditions. IH a combined footing 
is used for two unequal column loads, the base may be made in the 
shape of a trapezoid, the center of gravity of which coincides with the 
center of gravity of the column loads, 


REINFORCED CONCRETE FOOTINGS 023 


The parallel Sided: are placed at right are to the line joining 
the two column centers, the longer side of the trapezoid being at the 
heavier column. It is more desirable, however, to use a base of 
footing rectangular in shape, as shown in Fig. 173, p. 530, as the 
design and construction of a rectangular footing is simpler than that 
‘of a trapezoidal footing. The inequality of column loads is taken 
care of by extending the footing the required length beyond the 
heavier column. ‘The rectangular footing can be used only when the 
wall column load is equal to or smaller than the interior column load. 

With equal column loads, the combined footing should be sym- 
metrical when possible. Continuous footing for a row of equal col- 
umn loads should extend beyond the end columns, to provide 
sufficient bearing area on the soil for the same. 

The method of procedure in determining the base of facie is a8 
follows: The required area of the footing is computed. The center 
of gravity of the column loads is found. This point is then con- ’ 
sidered as the center of gravity of the rectangle or the trapezoid 
(whichever shape is selected) and the dimensions of the base are so 
selected as to give the required area, as previously computed. 

If the combined footing is for a wall column and an interior 
column, the footing at the wall column is made flush with the wall. 
The distance from the center of gravity of the column loads to the 
outside face of the wall is, therefore, fixed. This fixes the length 
of the rectangle, because the total length must be equal to twice this 
fixed distance from the center of gravity of the column loads to the 
outside face of the wall. When the length has been found, the width 
of the footing is determined by dividing the required area of the base 
by the length of the rectangle. 

If the combined footing is for two interior columns, the length of 
the rectangle is not fixed. Any number of combinations of width 
and length of footing are possible. The arrangement is most eco- 
nomical if the length of the longer cantilever, measured from out- 
side face of column, is equal to 0.35 of the net span between the 
columns. After ae length is determined, the width of the footing 
is computed from the required area. 

Design of Combined Footing.—In most cases, oe bending 
moments and external shears in a combined footing may be found 
by statics. After they are determined, the footing is designed by 
formulas and principles given in the chapter on Reinforced Concrete. 

The combined footing may be a slab of uniform thickness (Fig. 


524 FOUNDATIONS AND FOOTINGS 


173). If the thickness required for punching shear at the columns 
is larger than the depth required for bending moments and diagonal 
tension, a pedestal of proper depth and width may be placed over the 
slab at the column. When the shearing stresses permit it, the 
cost of the footing may be reduced by eliminating some of the 
concrete in tension and using an inverted T-section for the cross 
section of the footing. The bottom flanges of the T-section act 
as cantilevers. If the tensile stresses exceed the allowable stresses 
for plain concrete, sufficient reinforcement should be used in the 
flanges of the T-section to resist the bending moment. The whole 
section must be built in a continuous operation. 

Bending Moments.—Combined Footings for Two Columns.—A 
combined footing for two columns consists of either one or two 
cantilevers and a center span between the two columns. The 
footings are subjected to the reaction of the soil, which acts upward; 
hence, the bending moments are of opposite sign to the bending 
moments in ordinary beams where the loads act downward. On the 
cantilever, the bending moments are positive, producing tension in 
the lower part of the footing. In the middle portion of the center 
span, the bending moments are negative, producing tension near 
the top of the footing, while at the columns they are positive, 1.e., of 
same character as for the cantilevers. 


Let M, = maximum bending moment at the column, 
Ms = maximum negative bending moment in middle of center 
span; 
w = upward soil pressure, lb. per lin. ft.; 
1, = net span between columns in feet; 
l, = the length of cantilever in feet, 


I 


The maximum positive bending moments on the cantilevers, 
acting at the outside edge of the column or pedestal, are found by 
multiplying the upward soil pressure on each cantilever by the dis- 
tance from its center of gravity to the edge of the column or pedestal. 


Maximum Bending Moment for Cantilever. 
wl? . 
M, = rer ft.-lb. = 6wlo? in-Ib. . . . ~ (28) 
The positive bending moment at inside edge of the column (in the 
center span) is theoretically equal to the maximum bending moment 
in the adjoining cantilever. 





REINFORCED CONCRETE FOOTINGS 525 


The magnitude of the negative bending moment in the middle 
portion of the center span may be obtained by combining the nega- 
tive bending moments for the span, considered as simply supported, 
with the positive bending moments at the columns, produced by the 
cantilevers. When the bending moments at the two columns are 
equal (or nearly equal), the maximum bending moment acts in 
the middle of the center span and is equal to the maximum bending 
moment for a simple span minus the bending moment at the column. 


Maximum Bending Moment in Middle of Center Span. 
Then 


wl? 


Mz = M; — — ft-lb. Bs Say ey gs ge (26) 


For inch-pounds, multiply by 12. 

For a large difference between the cantilever bending moments at 
the two columns, the maximum bending moment in the center span 
will act off center, and the above rule will not apply. In such case, 
the maximum bending moment can be obtained by plotting the pos- 
itive and negative bending moments and determining the largest 
value by scaling or analytically, from the principle that the maximum 
bending moment in the center span acts at the point of zero shear. 
If the column load at the right is P1, then the point of zero shear is at 


a distance equal ton! from the right edge of the footing. Let Is be 


the distance of this point from the inside edge of the column, then 


2 
My = My ~ 2 feb. | seaman ba 


The above formulas are based on the assumption that the bending 
moments at both sides of the column are the same. This assumption 
is correct if the effect of the rigid connection between the column and 
the footing is not considered. Actually, this effect is appreciable 
enough to affect the bending moments in the center span. For 
cantilevers smaller than 0.3 of the net span the bending moment at 
the columns in the center span is larger than the bending moment 
produced by the cantilever. On the other hand, for long canti- 
levers, part of the bending moment due to the cantilever is absorbed 
by the rigidity of the column, so that only one part of it is transferred 
to the center span. 


526 FOUNDATIONS AND FOOTINGS 


To provide for this, the positive bending moment at the inside 
wl 


2 
———— ANI aon 


edge of a column with cantilever should not be less than 16 


Dy) 

wall column not less than ae Also, in the middle of the center 

span, sufficient reinforcement should be provided for a negative ~ 
wh? 

20.’ 

smaller, as would happen for cantilevers longer than 0.38/;. 


bending moment even if the computed bending moment is 


Combined Footing for a Row of Columns.—A combined footing for 
a row of three or more columns may be treated as a continuous beam. 
With properly proportioned footings, there is smaller chance than 
in floor construction of unequal reactions on the various spans due 
to the variation of the live load, because there is small chance for the 
aggregate area of the floor tributary to one column to be much more 
lightly loaded by live load than the aggregate area of the adjoining 
columns. Therefore, not as much provision for the unbalanced load- 
ing need be made as is customary in continuous beams, and the bend- 
ing moment in the center of spans may be smaller than in continuous 
beams in a floor. It may be made more nearly equal to the true 
theoretical moment with all spans loaded. It is a matter of judgment 
just how nearly equal the two bending moments can be made. An 
inexperienced designer is advised to use the same increased bending 
moment coefficients as recommended for continuous beams. 

If the spans are unequal, special computations should be made. 
The tables and diagrams given in Volume III may be used. 

External Shear. Footings for Two Columns.—Shears for com- 
puting diagonal tension and bond stresses are determined by statics. 
The shear from the cantilever at the edge of the column equals the 
total upward load on the cantilever. The shear at the edge of the 
column in the middle span equals the shear from the center span, if 
considered as simply supported, plus the shear due to the bending 
moment on the cantilevers. When the cantilevers on both sides are 
equal, the shears due to cantilever action balance; hence there is no 
increase in shear. For unequal cantilevers, the external shear at the 
edge of the columns with the larger cantilever will be increased by a 
value equal to the difference between the bending moments on the two 
cantilevers about the center of the columns, divided by the distance 
between the column centers. At the other column, the shear will 
be decreased by the same amount, The same results may be obtained 


REINFORCED CONCRETE FOOTINGS 527 


by subtracting at each end, from the column load, the upward reac- 
tion on the area between the edge of the footing and the inside edge 
of the column. 

External Shears for a Row of Columns.—The shears should be 
computed in the same manner as explained for continuous beams. 

Transverse Bending Moments.—If the width of the combined 
footing is larger than the width of the column, the upward pressure 
of the soil will produce a bending moment about an axis obtained by 
connecting the outside edges of the two columns. The part of the 
footing projecting beyond this axis acts as a cantilever, and the 
upward soil pressure produces a positive bending moment along 
the axis. This requires reinforcement placed near the bottom of 
the footing and running across the footing; this reinforeement may 
be either concentrated at the columns or spread along the whole width 
of the footing. While there is a preference for the former method, 
the latter method seems more logical. The transverse reinforce- 
ment is intended to prevent upward bending of the whole footing and 
not only of the portion near the column. The bending moment is 
produced all along the axis. It does not seem reasonable to resist 
it a great distance from the place where it was formed. The authors 
recommend, therefore, uniform spacing of the transverse bars over 
the whole footing. 

Diagonal Tension.—Diagonal tension stresses are likely to be 
large and should always be computed. Reinforcement should be 
used to provide for diagonal tension, when required by stresses. 
The external shears to be used may be computed as explained in the 
previous paragraph, under proper headings. The diagonal tension 
reinforcement is best provided for by means of stirrups, because, 
owing to the large depth of the footing relative to its length, it is 
seldom feasible to bend the longitudinal bars in such a way as to 
make them useful for shear and bending moment. 


Reinforcement for Combined Footing. 


Footing Carrying Two Columns.—The main reinforcement con- 
sists of longitudinal bars placed near the top of the footing in the 
center span and near the bottom at the two columns. The areas 
of steel are found from bending moments computed as explained on 
p. 524. 

At least 50 per cent of the steel in the center span should extend 
from column to column, The remaining bars may be bent down at 


528 - FOUNDATIONS AND FOOTINGS 


proper places and carried near the lower face of the footing into the 
cantilever to serve as reinforcement for the center span at the column 
and also as cantilever reinforcement. Sometimes, the depth of the 
footing is too large in proportion to the length, so that the bend of 
the bars would be too steep. In such cases, the bars may remain 
straight and extend about a foot beyond the points of inflection. 
Bars should be investigated for bond stresses. 

The steel at the columns consists of bent bars carried from the 
middle span, short bars, or a combination of both. Bond stresses 
are always very high and should be investigated. Although the 
bending moment is the same at both sides of the column, the shear 
is much larger at the inside edge. ‘This may require a much larger 
amount of steel than that required by bending moment. 

Cross reinforcement should be placed as explained under proper 
heading. 

Bond Stresses.—Bond stresses in footings are always high and 
should be computed. In the middle span for the top steel, the shear 
at the point of inflection should be used in Formula (50), p. 262. 
For the bottom steel at the column, the bond stresses should be com- 
puted at both sides of the column. At the column, the bending 
moment is the same on both sides, so that the number of bars required 
by bending moment is the same. 

Usually, however, the external shear at the inside edge of the 
column is larger than the external shear at its outside edge. Since 
the bond stresses depend upon shear, they are larger at the inside 
edge. The explanation is that the rate of increase in bending 
moment at the inside edge of the column is much greater and the 
tensile stresses in steel are developed in a much shorter distance 
than in the cantilever. In such cases, to reduce the bond stresses, 
the amount of steel must be increased, also the steel must be anchored 
at its free end to permit large allowable bond stresses. 

The example below illustrates the design of a combined footing. 


Example of Combined Footing. 


Example 6.—Find the dimensions of a combined footing for a wall column with 
a cross section 38 in. by 24 in., carrying a load P; = 400 000 Ib. and an interior 
column 30 in. square with a load P2 = 580 000 Ib. The distance between column 
centers is 15 feet and the allowable unit pressure on soil p = 8 000 Ib. per sq. ft. 
The outside edge of the foundation must coincide with the outside edge of the 
wall column. 





REINFORCED CONCRETE FOOTINGS 529 


Solution. Area of Footing.—Assume dead load of footing, W = 70000 lb., 
then total load on foundation, Pi + P2 + W = 400000 + 580000 + 70 000 = 
1050 000 lb. For allowable unit pressure, p = 8 000 lb., the required area of 

Pe 1-050:000 
footing is ee 131 sq. ft. 

Center of Gravity of Column Loads.—The distance of the center of gravity 
of the column loads from center of wall column is found by computing the moment 
of the column loads about the center of wall column and dividing it by the sum of 
the column loads. Dead load of footing needs no consideration in determining 
the center of gravity of the column loads. The column load Pz» is the only load 
producing a bending moment about the center of the wall column. This moment 
equals 15 X 580 000 ft.-lb. The sum of the column loads is 400 000 + 580 000 = 
980 000 Ib.; hence, the distance from center of wall column to center of gravity 


15 X 580 000 
ee = 8.9 it, 
980 000 


Shape of Footing.—Use a rectangular footing with one side flush with the 
edge of the wall column. Its length will be determined from the requirement 
that the center of gravity of the rectangle should coincide with the center of gravity 
of the loads. 

The distance from the edge of the footing to the center of the wall column is 
1 foot. Since the distance from center of wall column to center of gravity of the 
loads, as computed above, is 8.9 ft., the distance from the edge of the footing to 
the center of gravity equals 1 + 8.9 = 9.9 ft. The center of gravity of a rect- 
angle is in its middle. Therefore, 9.9 ft. found above, equals one-half of the 
length of the rectangle. Its full length, therefore, is 9.9 K 2 = 19.8 ft. 

Width of Footing.—Since required area of footing is 131 sq. ft., and its length 


131 
19.8 ft., the width equals Tee 6.6 ft. Use 6 ft. 9 in. 


Unit Pressure for External Shears and Bending Moments.—Bending moments 
and external shears are produced by the column loads only, exclusive of the dead 
load of the footing. 

400 000 + 580 000 


Unit pressure per sg. ft. is —79.8X6.75 = 7 340 lb. 


400 000 + 580 000 
Pressure per lineal foot of footing arta = 49 500 lb. 


Bending Moments.—As evident from Fig. 173, p. 530, the footing consists of 
the center span and of the cantilever beyond load P». 

Bending Moment of Cantilever—Length of cantilever to the edge of the out- 
side column is 2.55ft. Since pressure per lineal foot equals 49 500 lb., the bend- 
ing moment on the cantilever is total load multiplied by moment arm, or 


2.55 : 
M, = 49 500 X 2.55 X = < 12 = 1930 000 in.-lb. 


This moment is positive, causing tension at the bottom face of the footing. 


530 FOUNDATIONS AND FOOTINGS 


Bending Moment in Center Span.—At the inside face of the interior column, 
the bending moment will be either the same as produced by the cantilever or 


1,2 eee 
ae ft. lb., whichever is larger (where w = pressure per lineal foot and 1; 


distance in feet between inner faces of columns). For w = 49 500 lb. h = 


wl 12 


[15 — (1.0 + 1.25)] = 12.75 ft., the bending moment from formula Ae is 
12.75? 
pies et IEE ius ~ = 503 000 ft.-lb., or 6 036 000 in.-lb. This moment is larger 


than the bending moment due to the cantilever and will be used in computing 
the required reinforcement at the column. 








Elevation 


"Ra, Stirrups 12 legs each 





, 


Plan 
Fic. 173.—Combined Footings. (See p. 529.) 


In the middle portion, the maximum bending moment will be computed at 
the point of zero shear. The distance of the point of zero shear from the outside 
edge of the wall column equals the wall column load divided by the pressure per 

400 000 

49 500 
ft. from inside edge of column. The maximum negative bending moment is now 
found by statics. Disregarding the indefinite amount of restraint at the wall 
column, the negative moment, from Formula (27), p. 525, where M, = 0 and 
w = 49 500 lb. and J; = 6.1 ft., equals 


lineal foot. Itis 





— 8 1 ft. from outside face of column, or 8.1 — 2 = 6.1 


6.1? 
M, = — 49500 X ih gant ak 921 000 ft.-Ib. or — 11 052 000 in.-lb. 





REINFORCED CONCRETE FOOTINGS 531 


External Shears. 

Shear Due to Cantilever.—External shear due to the cantilever at the outside 
edge of interior column, is equal to the pressure per unit of length multiplied by - 
the length of the cantilever in feet. It acts downward and is therefore negative. 


Vi = 49 500 X 2.55 = 126 200 lb. 


Shears in Center Span.—The external shears at the inside edge of the columns 


will be computed by subtracting the upward pressures from the respective col- 
umn loads. 


External Shear at Inside Edge of Interior Column. 

V2 = 580 000 — 49 500 X 5.05 = 330 000 lb. 
External Shear at Inside Edge of Wall Column 

Vs; = 400 000 — 49 500 X 2 = 301 000 lb. 


Depth of Footing.—Depth of footing is determined either by bending moment 
or by diagonal tension, whichever gives the larger value. The depth should be 
checked for punching shear and, if required, a pedestal should be added on the 
top of the footing. 

Depth Determined by Bending Moment.—In this case, the largest bending 
moment is the maximum negative moment of the center span, M = 11 052 000 


| M 
in.-lb. Using Formula (1), p. 204, d = C . ae in which width of footing b = 81 


in. and for unit stresses f, = 650 lb. per sq. in., fs = 16 000 lb. per sq. in. and 
n = 15, C = 0.096 (see p. 880). Depth as required by bending moment 


11 052 000 . 
d = 0.096 (ee = 5055 In. 


Diagonal Tension.—The largest external shear is at the inside edge of the 
interior column. For the depth found above, d = 35.5 in., and a width of foot- 


Vv 
ing equal to 6 ft. 9 in. or 81 in., the unit shear from formula v = bid? is 
J 
330 000 — 


a BA 0.875 X 85.5 





= 131 lb. per sq. in. 


This value is larger than the allowable working limit, which is 120 lb. per sq. in. 
The depth as determined for bending moment must be increased in the proportion 
of 130 to 120. The depth as required by diagonal tension is d = 35.5 X 434 
= 38.7 in. Diagonal tension reinforcement must be provided for sections 
where v exceeds 40 Ib. per sq. in. 

Longitudinal Reinforcement.—Using depth of slab d = 38.7 in. as required 
by diagonal tension, the amount of steel at the various sections will be found 
from equation 

M 


maga 


532 FOUNDATIONS AND FOOTINGS 


where j = 0.875 and f, = 16 000 lb. per sq. in. 


Center of Footing, 
M, = 11 052 000 in.-lb. 
A, = 19.5 sq. in: 
Twenty-six $-in. bars will be tried. 


Center Span at Interior Column, 
M = 6 036 000 in.-lb. 
Age=" 11 2-s05107 
Twenty-six 3-in. round bars will be tried. 


The bars must be extended into the cantilever. 
Cantilever, 

M, = 1930 000 in.-lb. 

A; = 3.6 8q. In. 


This amount of steel is smaller than the amount extended from the other side 
of the column. 

Bond Stresses.—It is necessary to compute bond stresses in top and bottom 
reinforcement, as the footing may fail by slipping of the top or bottom bars. 

Bond Stresses in Top Reinforcement.—Since, theoretically, there is no bending 
moment at the wall column, the bond stresses in the top steel there will be com- 
puted as for simply supported beams (note that top steel in footings corresponds 
to bottom steel in ordinary beams). At the other end of the footing, the bond 
stresses are smaller, because there they have to be computed at the point of 
inflection where the external shear is smaller. The shear at the edge of wall 
column is V; = 301 000 lb. The bond stress with twenty-six {-in. square bars, 


having a total perimeter 20 = 26 X'3.D.= Olam from formula wu = Soja’ iS 
0 
301 000 08 1b : 
“= : meh. 
91 X EX 38.5 Pe 


This is satisfactory for deformed bars. 

Bond Stresses in Bottom Steel—The amount of bottom steel is the same at 
both sides of the interior column. The external shear, however, is larger at the 
inside edge. The bond stresses will be figured for the larger shear. 

The perimeter of twenty-six 4-In. Sq. bars is 61.2 in. For Vz = 330000 lb. 
and depth 38.7 in., the bond stresses are 


330.0000) 100 aa 
= = . per sd. 4 
R198 SPO EST Be ae 





U 


To reduce the excessive bond stresses, use twenty-nine 5_in. sq. bars, the perim- 
eter of which is 29 X 2.5 = 72.5 in. 


330.000 
72.BI toe o0.t 





uU = 134: 


Further reduction of bond stresses by reducing the diameter of bars is not advis- 
able. Instead, the depth of the footing may be increased by providing a block 


<< 


STRAP BEAMS TO CONNECT FOOTINGS 0933 


at the column, and the bars will be extended beyond the point of inflection, in 
which case larger bond stresses are allowed. 

Transverse Reinforcement.—The projection of the footing beyond the edge 
of the interior column is 2 ft. 13 in., and beyond the edge of exterior column 
1 ft. 93 in. The transverse bending moment will be figured for both projections. 
The upward pressure per square foot is 7 340 lb. Hence, transverse bending 
moments per foot of length of the footing are: 


een 
M; = 7340 X 2.12 X eee 198 000 in.-lb. 


21.5 
_M, = 7340 X 1.79 x oe 141 000 in.-lb. 


For d = 38.5in. f, = 16000 





198 000 0.37 : 
Ss ae ae y Tepe Gin al . sq. 1nN. 
> 16 000 X $ X 38.5 ae 
Use 3-in. rd. bars, 10 in. on centers. 
142 000 : 
A sq = 0.26 sq. in. 


~ 16000 X 2 X 38.5 


Use 3-in. rd. bars, 14 in. on centers. 


STRAP BEAMS TO CONNECT FOOTINGS 


Instead of using a combined footing for a wall column and an 
interior column, as in the previous discussion, it may be desirable 
to build both footings separately and to take care of the eccentricity 
on the wall footing by connecting it with the adjoining interior foot- 
ings by means of strap beam. ‘The strap beam may or may not be 
used to transmit the load to the foundation, depending upon the 
design. Its primary function is to resist the bending moment pro- 
duced by the eccentricity of the wall column load with respect to the 
wall footing reaction. 

Such an arrangement is particularly useful when pile foundation 
is used for both columns, and it is desirable to place the piles in 
clusters next to the columns; it is also useful when the footings for 
both columns consist of caisson piles. 

Bending Moments and Shear in Strap Beams.—For the purpose 
of determining bending moments and shears, the strap beam may be 
considered as a balanced double cantilever supported at the center of 
the wall footing. The short cantilever carries the wall column load, 
P,, and the long cantilever carries the balancing reaction, which is 


534 FOUNDATIONS AND FOOTINGS 


provided by the interior column with which the long arm is con- 
nected. The conditions of loading are seen from Fig. 174, p. 534. 

To make the bending moments statically determinate, the strap 
beam is considered as free to rotate at the ends. With this assump- 
tion, from simple mechanics, the bending moment is zero at both 
ends and increases to a maximum at the point of support. 


_--For Free End of Long Arm 









P,=600,000 tb 


~ 
ek 


Loading Diagram | 






5-1" 8q. Bars 
Caisson---- 











0 ee 


Design of Strap 





Fic. 174.—Loading and Moments in Straps. (See p. 534.) 


Let Pi = load on wall column, |lb.; 
1 = distance between centers of the wall footing and the 


interior footing, ft.; 
e = distance between center of wall column and wall foot- 


re, tee 


R, = reaction on wall footing, lb.; | 
Rs = reaction on interior footing due to load P, lb.; 
M = bending moment. 


STRAP BEAMS TO CONNECT FOOTINGS 535 


Then 
Maximum Bending Moment in Strap Beam. 
ese eae il, oe as es has) 
Reaction on Wall Footing. 
R, = P{1 +5) 1 Sar ous Sie a 20) 
Reaction on Interior Footing due ta Load P. 
Ro = — Pi, Ib. Mee eee ese La} 










Dowels 





Point of Zero Shear 


NV 
‘- Stirrups 





Elevation 


Fic. 175.—Strap for Footing with Uniformly Distributed Reaction. (See p. 536.) 


It should be noted that the load for which the wall footing is 
designed is larger than the wall column load. The load on the interior 
footing is decreased by the reaction Ro. 

The bending moment in Formula (28) is based on the assump- 
tion that the reaction R, is concentrated in one point. This assump- 
tion gives the maximum possible bending moment. Actually, the 
reaction is distributed and.the bending moment is therefore smaller. 

Two cases may occur: (1) the reaction A; consists of uniformly 


536 FOUNDATIONS AND FOOTINGS 


distributed upward pressure; (2) the reaction consists of several 
rows of piles. In both cases the maximum bending moment acts at 
the point of zero shear. 

For uniformly distributed reactions, the distance of the point of 
zero shear from edge of footing is found by dividing the column load 
P, by the reaction per lineal foot. Such a condition is shown in 
Fig. 175. Assume that the unit reaction of foundation is p and that 
P = pag; then ag is the location of the point of zero shear from edge 
of footing reactions. If a; is the distance between the load P and 
the point of zero shear, then 


M = —Pi(a—$). . Ree aS Lee 





Fic. 176.—Straps for Footings on Piles. (See p. 536.) 


For footings supported on piles, the location of maximum bending 
moment is at point of zero shear or point where shear changes sign. 
Usually this will be at the center of the last row or piles. If a is 
the distance from this point to the load Pi, qi is the rezstion of the 
first row of piles and its distance a2, gz the reaction of the second row 
of piles and its distance a3, then 


= — [Pya, — (qid2 + qoaa)].. . . - « (82) 


For a larger number of piles it may be necessary to use a3, a4, etc. 

Dead Load of Footing and Strap Beam.—The dead load of footing 
and strap beam should be added to the footing load, f1, in computing 
its size. In figuring the bending moment of the strap beam, however, 
the dead load of the strap may be disregarded. 


STRAP BEAMS TO CONNECT FOOTINGS 507 


Design of the Strap Beam.—The wall footing is designed first, 
and its center fixed. This gives the value of e and l. With these, 
the bending moment is determined. The width of the strap beam 
is made the same as the width of the column. When the bending 
moment and the width are known, the depth is computed from 


Formula (1), p. 204 (a = ales) 


For the selected depth, the area of steel is computed from 
Pe 

jaf 
The depth required by the bending moment is usually larger than 
that required by diagonal tension. It should be noted that the 
external shear producing diagonal tension in the long cantilever is 


A, 


constant and is equal to P be 


Since the bending moment decreases toward the interior column, 
the depth of the strap beam may also be decreased. The minimum 
depth, of course, may not be smaller than required by diagonal 


tension produced by the shear Bir 


Reverse Bending Moment at Interior Column.—In the above 
discussion, it was assumed that the end of the cantilever at the 
interior column is free to turn. Since the strap beam must be 
connected either with the footing or with the column, this assump- 
tion is not correct. Owing to the rigidity of the connection of the 
strap beam with the interior column, some positive bending moment 
will be developed in the strap beam at the column. If the strap 
beam is fixed at the interior column, this bending moment is equal 
to one-half of the maximum bending moment, Pe. For intermediate 
conditions, intermediate values may be assumed. It is advisable 
to make generous provision for this bending moment. 

Provision for Unequal Settlement.—The two footings joined by 
a strap beam may settle unevenly. This may produce stresses of 
opposite character to the stresses for which the footing is designed. 
To provide for this, the strap beam should be reinforced on top and 
bottom for its whole length. The authors recommend the use of 
one-third the amount of the steel in the bottom. This recommen- 
dation is arbitrary. 

Anchoring Reinforcement at Wall Column.—In the short canti- 
lever, full tensile stresses in the kars must ke developed in the short 


538 FOUNDATIONS AND FOOTINGS 


distance between the center of column and the center of footing. 
This distance usually is not sufficient for the purpose. The reinforce- 
ment must be extended to the edge of the wall column and then pro- 
vided with a hook sufficient to develop the tensile stresses in the bar. 
Without such provision the construction would fail by bond. 

Types of Strap Beams.—F'igures 174 to 176 illustrate several of 
the types of strap beams most commonly used. 


Example of Strap Beam Design.—In Fig. 174, p. 534, the columns rest on 
caissons. 
The data are as follows: 


P, = 600 000 lb.; 

P, = 1000 000 Ib.; 

16 in. = 1.33 ft.; 
20 ft., 


® 
I 


I 


€ 


1.33 
R, = Pa(i Bi | = 600 000 ( Ae) = 640 000 lb.; 


R, = Py = 40 000 Ib.; 


S 
T 


Pe = 9600 000 in.-lb. ; 
b= 44in. d= 45in. A, = 15.3 sq. in. 


The arrangement of reinforcement is evident from the figure. 


RAFT FOUNDATION 


When the allowable pressure on the soil is small, it may be neces- 
sary to spread the foundation over the whole area of the building. 
Such a foundation is called a raft foundation. It is often used in 
connection with piles, where they are driven in comparatively soft . 
strata and it is desirable to space them as far apart as possible so as 
not to overload the ground in which the piles have bearing. ‘The 
raft foundation is made either of flat slab construction or of beam and 
slab construction. | | 

Pressure on Foundation—As in the case of independent foot- 
ings, to prevent unequal settlement, it is necessary to design the 
foundation so that the unit pressure on the soil is uniform. Tor 
uniformly distributed floor loading, this is not difficult. If the area 
of the foundation is equal to the area of the building, then the load 
on the foundation will be equal to the sum of the loads on all floors. 


RAFT FOUNDATION 539 


With uniformly distributed floor load, the foundation load will also 
be uniform. 

The difficulty begins when the floor loads are not uniform, or 
when some columns carry additional loads. (Columns carrying the 
tank, for instance.) At wall columns also, the loading is proportion- 
ately larger than at the interior columns because it consists not only 
of the floor loading, but also of the weight of the wall. To provide 
for the additional load, it is necessary to extend the foundation out- 
side of the building lines. The additional area to be provided is 
equal to the excess load on the wall column divided by the unit pres- 
sure on the foundation. When it is not possible to extend the 
foundation outside of the building, uneven pressure on the founda- 
tion will result. When this increased pressure does not exceed the 
maximum allowable pressure, the raft foundation may be used; but 
the construction between the wall columns and the first row of 
interior columns should be made strong enough not only to resist 
the upward pressures, but also to even up the settlement of the two 
columns. Uneven settlement of the wall column produces negative 
bending moment at the interior column and positive bending moment 
at the wall column. For large differences in unit pressure produced 
by the heavy columns, the design may be equalized by using piles 
under the heavy Arias 

If raft foundation is used in connection with piles, the TUN 
in column loading may be taken care of by using closer spacing of 
piles at the heavier loaded columns. 

Flat Slab Foundation.—The foundation may be of flat slab design, 
consisting of flat slab, drop panels, and column heads, as described 
in the chapter on Design of Flat Slab Structures, p. 319. The same 
formulas may be used as recommended for flat slab floors. 

In this case, the pressure on the flat slab acts upward and the 
column load downward. The loading is of opposite sign to the load- 
ing in a floor; therefore the bending moments are also of opposite 
sign. At the columns, the bending moments are positive (instead 
of negative), producing tensile stresses at the bottom; therefore the 
tensile steel must be placed at the bottom of the slab. In the 
central parts of the slab, the bending moments are negative (instead 
of positive, as in a floor slab) and the reinforcement must be placed 
near the top of the slab. 

To save form work, the drop panel may be placed below the slab. 
Instead of a column head of conical design, a cylindrical block of 


540 FOUNDATIONS AND FOOTINGS 


concrete of a diameter equal to the diameter of the column head may 
be used. This is placed on the top of the slab. 

Cinder fill, equal in depth to the height of the column head, is 
provided on the top of the foundation slab. 





Fic. 177.—Details.of Flat Slab Foundations. (See p. 539.) 


Beam and Slab Foundation.—The design of the foundation may 
consist of a combination of beams, girders, and slabs. ‘The principles 
of floor design must be followed in determining the concrete dimen- 
sions and the amount of steel. As explained in the previous case, 
the loading acts in the opposite direction to the floor loading, so that 
the location of the steel will be opposite to that in the floor, 





RAFT FOUNDATION 541 


The beams, or ribs, may be built either above or below the slab. 
When the ribs are above the slab, the beams are T-beams in the 
center and rectangular beams at the support. With ribs placed 
below the slab, the beams are T-beams at the columns and rectangu- 
lar beams in the center of the span. 

The design with beams above the slab is preferable, as better 
bearing on foundation is obtained. The space between the ribs is 
filled with cinders and the basement slab is placed on the top. It is 
necessary that the ribs be built monolithic with the slab. There is a 
tendency to pour the slab first, place the forms for the ribs on the 
slab, and finish pouring the ribs. This is absolutely wrong, as it is — 
impossible to make the ribs and the slab act as a unit when they 
are poured in this way. 


CHAPTER [IX 
PILES 


A foundation may be supported on piles when the ground, for a 
ereat depth, consists of layers of soft materials on which it is either 
impossible or undesirable to rest the structure. The piles may be 
driven either to solid bearing on rock or hardpan or far enough into 
the firm ground to develop, by frictional resistance, the allowable 
capacity of the pile. 

Piles may be of wood or concrete. The choice will depend upon 
economy and also upon the conditions affecting the permanency of 
the structure. For instance, when there is danger of their being 
attacked by teredo, or a likelihood that the level of low water will be 
materially lowered by future drainage, concrete piles should be given 
preference even at higher cost. The two kinds of piles will be treated 
separately. 

Spacing of Piles—The minimum spacing of piles depends upon 
two conditions. First, the piles must not be driven close enough 
together to compress the ground to an extent which would make 
impossible straight driving of adjacent piles. Second, the spacing 
must be such that the load transferred to the soil in which the pile 
has its bearing will not exceed the bearing value allowable on such 
soil. For instance, if the pile is driven into clay with an allowable 
unit bearing of 4 tons and the capacity of the pile is 20 tons, the area 
tributary to this pile must be at least 5 sq. ft., which requires a 
minimum spacing on centers of about 27 in. If piles are driven too 


close together the whole cluster with the surrounding earth may sink — 


as a unit. 

The minimum spacing on centers, as specified by most codes, is 
from 20 to 24 in. This spacing may be used if the ground is not 
compressible. New York City specifies a minimum spacing of 
20 in., and a maximum of 36 in. In railroad work, closer spacing 
than 36 in. is not permitted. For use in building construction, 
under ordinary conditions, the authors recommend a spacing of 
piles of 30 in. on centers. 

| 542 


WOOD PILES 543 


Piles for Outside Bearing Walls.—All bearing walls should have 
at least two rows of piles. or buildings over three stories in height, 
if two rows of piles are used, the spacing of the rows should be at 
least 36 in. on centers. . 


WOOD PILES 


Wood piles are commonly used for foundations because of their 
low cost. They may be either of soft timber, such as pine, spruce, 
or hemlock, or of hard timber, such as oak, cedar, or hickory. The 
following are the requirements for wood piles: They shall be sound, 
close-grained, solid, and without injurious ring shakes or loose knots. 
All knots shall be trimmed close to the body and all bark peeled 
soon after cutting. Piles shall be of uniform taper; not less than 
6 in. in diameter at the point, and not less than 10 in. at the butt, 
for piles up to 25 ft. in length; and not less than 12 in. at the butt 
for longer piles. ‘They shall be practically straight, a criterion of 
straightness being that a line drawn from the center of the butt to 
the center of the point shall be within the body of the pile. Short 
bends shall not be allowed. 

The carrying capacity of the wood pile depends upon the char- 
acter of the material into which it is driven. If the pile is driven to 
solid bearing on rock or hardpan, through firm material which is 
able to keep the pile from lateral displacement, its bearing capacity 
is equal to its strength as a short column. In such cases, the authors 
recommend 20 tons as the capacity of a pile. This capacity must be 
reduced when the pile is driven through material which, for any 
appreciable distance, does not offer resistance to lateral displace- 
ment of the pile. Such materials are water, mud, silt, peat, or fill. 
The pile should then be figured as a long column of a height equal to 
its unsupported length and of an average cross section within this 
length. 

If the pile is not driven to solid bearing, its capacity depends upon 
the frictional resistance in bearing soil. It transfers the load to the 
soil gradually by friction. The capacity then is best determined by 
a load test. Where this is impossible, the capacity of the pile, 
i1.e., the load to be used in design, is ascertained from its penetration 
under the last blows of the hammer. The most common formulas 
for the capacity of piles are the Wellington formulas, also known as 
the Engineering News formulas, given below. Their use is recom- 
mended by the authors. : 


544 PILES 


P = capacity of pile, lb.; 
W = weight of hammer, lb.; 
h = fall of hammer, ft.; 
s = penetration of pile at last blow, in., or average pene- 
tration for several consecutive blows. 
2Wh 


Then P = ear for pile driven with drop hammer, . . - (1) 





2Wh : 
P= ea for pile driven with steam hammer... . (2) 





Modern specifications require that s be taken as the average 
penetration for a series of blows, usually five to ten, and further 
require that the pile should be driven until the successive blows 
produce approximately equal or uniformly diminishing penetra- 
tions. | 

When the piles are designed for a certain capacity, they are 
driven until the final penetration, s, becomes small enough to give 
the required capacity when computed by Formulas (1)° or (2) 
The weight of the drop hammer and the drop being known the max- 
imum allowable penetration is computed and the pile driven until 
the penetration is reduced to this value. For instance, if a pile is 
designed to carry 20 tons, the weight of the drop hammer is 4 000 lb. 
and the drop is 10 ft., the maximum allowable penetration giving the 
required capacity obtained from the formula is s = one inch. The 
pile is driven until the penetration is reduced to one inch. 

If it is impossible, with the length of pile at one’s disposal, to get 
small enough penetration at the last blows, the capacity of the pile 
should be reduced according to the value of obtained final penetra- 
tion and the proper number of piles added to give the desired resist- 
ance to the foundation. The size and strength of the footings should 
be increased if required by the larger number of piles. 

Cutting of Wood Piles. All wood piles must be cut below perma- 
nent low water, so that the entire pile will always be wet. In such a 
case, unless destroyed by teredo, the piles may be expected to last 
indefinitely. It is important to determine whether there is any 
likelihood of the lowering of the low water level through future 
drainage. If so, the cutting level must be established with this in 
mind. The pile must be cut ‘to sound wood which has not been 
injured by the hammer, as otherwise its value would be negligible. 


WOOD PILES 545 


Driving Piles.—Piles should be driven under the direction of men 
experienced in this work. They should be driven straight, since the 
value of inclined piles for vertical loads is small. Some engineers 
require that piles driven to hard strata be provided with metal shoes 
of proper design. The depth to which piles are to be driven should be 
determined by borings, in order to prevent overdriving of piles, 
which is even more harmful than underdriving. An overdriven 
pile is one that has been forced farther down after the point has 
reached solid bearing. The result is that it buckles, breaks, or splits, 
at the end. In any case its value is either lost or greatly reduced. 
If possible, all the piles in a cluster should be driven to the same 
depth. 

Piles are driven by a drop hammer or a steam hammer. To 
prevent splitting of the tops, cushions are used. <A water jet is often 
used in connection with the driving of piles, when it is possible to 
make the soil flow. 

Weight and Fall of Drop Hammer.—The term drop hammer is 
used when the hammer is raised and then allowed to fall freely for 
10 to 20 ft. before striking the pile. The raising is usually done by a 
hoisting drum. 

The required weight of drop hammer depends upon the char- 
acter of the soil and the weight of the pile. For hard soils, the 
hammer must be heavier than for soft. It should also be heavier 
than the pile, as otherwise a large part of the effect of the blow will 
be used up in overcoming the inertia of the pile. In general practice, 
the weight of hammer for wood piles varies from 2 000 to 5 000 Ib., 
the most common weight being around 3 300 lb., and the fall from 
5 to 20 ft. The best results are obtained when a heavy hammer is 
used with a comparatively low fall, as then the brooming of the butt 
is prevented. Another advantage of low fall is that the succession 
of blows is more rapid, thereby preventing the set of the soil around 
the pile. 

Steam Hammer.—<As the name implies, the steam hammer is 
operated by steam. It may be either single-acting or double- 
acting. A single-acting hammer is lifted by steam and falls by 
gravity, while for the double-acting hammer steam is used both in 
raising and in the fall. The weight of the striking part varies for 
different makes of hammers. For single-acting hammers, the 
weight of the moving part is around 5000 lb., the stroke (or fall) 
around 3 ft., and the number of blows per minute about 60. The 


546 PILES 


weight of the moving part ‘na double-acting hammer is much smaller, 
varying from 1 250 to 2 550 lb. ‘The downward pressure, however, 
of the weight and the steam is around 7000 lb. The number of 
blows per minute varies from 100 to 200, the larger number being 
used with lighter hammers. 


CONCRETE PILES 


Concrete piles are widely used. They have the following 
advantages over wood piles: (1) They have larger carrying capacity 
per pile, thus reducing the number of piles and decreasing the size 
of the foundation; (2) concrete piles do not need to be cut below low 
water, which often results in saving of expensive excavation; (3) 
they may be built of any length and size desired; (4) concrete piles 
are less affected by the teredo and other wood borers. (5) Concrete 
piles are immune to decay. 

Concrete piles may be divided into three general groups: (a) pre- 
cast piles; (b) piles cast in place; (c) caisson piles. 

Pre-cast Piles.—As the name implies, these piles are cast in the 
yard before the construction is begun, and then driven in the same 
manner as wood piles. Pre-cast piles, being simply long reinforced 
concrete columns, are designed according to the regular principles 
for reinforced concrete. There are several patented pre-cast piles 
on the market, but the patented features apply either to the method 
of driving, the design of the top of the pile to prevent it from split- 
ting during driving, or the design of the shoe. 

Description of Pre-cast Piles.—The cross section of the piles may 
be square, round, or octagonal. Square piles with chamfered 
corners are commonly used. The sides of the piles are preferably 
made straight. . 

Some designers prefer tapered sides because they are easier to 
drive, and also because some experiments tend to show that for 
piles relying on friction- tapered sides are preferable. No taper 
should be used for piles that are to be driven into incompressible 
plastic clays, because such soils heave up, instead of compressing, 
when piles are driven into them. ‘There is thus a tendency for the 
piles that are already in place to be heaved up when adjacent piles 
are driven. 

When necessary, because of the character of the soil, the piles 
may be provided with shoes to facilitate driving and to prevent the 
end of the pile from breaking when it strikes a boulder. 


CONCRETE PILES 547 


The size of the cross section of the pile and its length are gov- 
erned only by practical considerations. That cross section which 
will best suit the conditions,.particularly with regard to ease in hand- 
ling and driving, should be selected. As the handling and driving 
of long piles having large cross section may require costly equipment 
and heavy hammers, it is often preferable to use a larger number of 
piles of smaller cross section. . | 

Capacity of Pre-cast Pile—Iif driven to solid bearing, the pile 
acts as a column and its capacity is equal to its compressive strength, 
as discussed on p. 406. 3 

If the carrying capacity of the pile is developed by frictional 
resistance, its value is determined either by load tests or by a formula 
from the penetration at the last series of blows. Most of the build- 
ing codes require a load test for concrete piles, as the formulas used 
for wood piles are not directly applicable. 

Mr. Charles R. Gow suggests the following modification of the 
Wellington formulas for pre-cast concrete piles: 


Let P = capacity of piles, lb.; 
W = weight of hammer, lb.; 
h = fall of hammer, ft.; 
s = penetration of last blow (or average of several blows), 





in. ; 
w = weight of pile, lb. 
Then P = Hell for drop Teel, aaa aes nacre UO) 
WwW ; 
Sa 
Ww 
and P= ES: SOUL WAMINIGT Seam (lets aoe Pe oe Ce) 
Ce S oee nee 
10w 


These formulas have not been as thoroughly checked as the formulas 
for wood piles, upon which they are based, but are recommended as 
the best present practice. They take into account the fact that, 
because of the greater mass of the pile, some of the force of the blow 
is lost in overcoming the inertia of the pile and is not available for 
producing penetration. 

Stresses in Pre-cast Piles.—The pile must be designed to resist 
(1) the stresses developed in the pile after it is in place, (2) the 
stresses during handling, and (3) the stresses during driving. 


548 PILES 


After the pile is in place, it acts like a reinforced concrete column 
of an unsupported length depending upon the character of the soil 
through which it was driven. If driven through firm soil offering 
sufficient resistance against lateral movement, the pile acts like a 
short column. If driven through fill, silt, peat, mud, or water, 
which do not offer any lateral support, the pile acts as a long column 
and its unsupported length is equal to the depth of such material. 
The allowable stresses on the pile in all cases should be the same 
as for a reinforced concrete column of the same design and with the 
same ratio of slenderness. Its strength should be figured by rein- 
forced concrete column formulas. (See Formulas (4) to (11), 
p. 406.) Piles driven to solid bearing on rock or hardpan should 
have straight sides, as the required strength is the same at all sec- 
tions. In frictional piles, the load carried by the pile is gradually 
transferred to the ground, so that the lower sections are subjected 
only to a fraction of the total load on the pile. This makes it pos- 
sible to taper the pile. 

Stresses During Handling.—The pile is usually poured in hori- 
zontal position. While being lifted by a derrick it is subjected to 
bending produced by its own weight. The character and the mag- 
nitude of the bending depend upon the number of points of suspen- 
sion. Piles up to 40 ft. in length are often suspended at their center 
of gravity during handling, so that the two parts of the pile act as 
cantilevers, producing maximum bending moments at the point of 
suspension. ‘Tension and compression stresses are a maximum at 
the point of suspension. Long piles suspended in the middle would 
require a large amount of steel to resist bending stresses. Therefore, 
for lengths over 40 ft. they should be suspended at two points, 
during handling, while for very long piles three points of suspension 
may be necessary to reduce bending stresses. It is clear that the 
designer of the pile must know how the pile will be handled, in order 
to provide proper strength for bending. 

As a much smaller factor of safety is required in this case, the 
allowable unit stresses may be 25 per cent larger than those recom- 
mended for permanent stresses. 

Pile Suspended at One Point. 


Example 1.—Find the required amount of steel for a 20-in. square pile of uni- 
form cross section, 40 ft. long. Pile to be suspended in the middle in handling. 


202 
Solution —The weight of the pile is 144 x 150 = 417 lb. per lineal foot. 


CONCRETE PILES 549 


Since the pile is suspended in the middle, the span of the cantilever is 20 ft. 
With a load of 418 lb. per lin. ft., the moment at the point of suspension is 


M = 417 X 20 X 10 X 12 = 1 000 000 in.-lb. 


With steel placed 2 in. from the surface of the pile in the clear, the effect- 
ive depth of a 20-in. section is 17.5 in. If the allowable stress in steel, f, = 
16 000 plus 25 per cent = 20000 lb. per sq. in., the required amount of steel, 


from f la A Le 
rom formula A, = —-, 1s 
jdf 
‘ 1 000 000 
~ 0.89 X 17.5 X 20000 


Ss 


= 3.2 sq. in. 
Four 1-in. round bars near the tensile face will be used. 


If the tensile face is definitely known, as is the case when pro- 
vision is made for attaching the lifting hook, it is sufficient to provide 
this reinforcement in the tensile face only; otherwise, the same 
amount of steel should be used near each face. The corner bars 
may be counted at two faces. 

Pile Suspended at Two Points.—When a pile with straight sides is 
suspended at two points in handling, the points of suspension should 
be located a distance equal to 0.206 of the pile length from each end 
of the pile. The bending moments at the points of suspension are 
then equal to the bending moment in the center of the pile. Such an 
arrangement requires the minimum amount of steel. 


Example 2.—Find the amount of steel to resist bending stresses in handling 
for a 20-in. square pile of uniform section, 80 ft. long. Pile to be suspended at 
two points in handling. 

Solution—The points of suspension will be placed a distance from the ends 
equal to 0.2061. = 0.206 x 80 = 16.5 ft. The length of cantilevers, therefore, 
will be 16.5 ft. and the length of the middle span 80 — 16.5 X 2 = 47.0 ft. 


202 
Weight of pile equals 44 X 150 = 417 lb. per lin. ft. 


Bending moment in cantilever at points of suspension, 
16.5 
M, = 417 X 16.5 X ek xX 12 = — 682 000 in.-lb. 


Bending moment in the center of middle span, 
M =: 417 X 47? x 12 — 682 000 = 1 390 000 — 682 000 = 708 000 in.-lb. 


From the above figures, it is evident that for the accepted spacing of suspension 

points the bending moments at the three critical sections are practically equal. 
With clear distance from steel to surface of pile equal to 2 in., the effective 

depth of a 20-in. section is d = 17.5 in. and the required area of steel for the 


550 PILES 


largest bending moment, using fs; = 20000 lb. per sq. in., from formula A; = 
M 





aS is 
jaf 8 
708 000 
A, = eB ie 
0.89 X 17.5 X 20 000 
Use four 2-in. round bars. 
2.27 
Since the ratio of steel is p = 17.5 X20 = ().0065, the compression stresses 


are satisfactory. The advantage of suspending the pile at two points during 
handling is evident from the fact that the 80-ft. pile requires less steel for bending 
than the 40-ft. pile of the previous example, which was suspended in the middle. 


Stresses in Pre-cast Pile During Driving.—Stresses during 
driving are those induced by the blows of the hammer at the top and 
those occasioned by uneven driving. Since concrete is brittle, 
direct blows of the hammer during heavy driving are apt to fracture 
and split the top of the pile unless it is protected by a cushion or the 
head is reinforced by special lateral reinforcement. Cushions reduce 
the effectiveness of the blow and slow down the progress of the 
driving, hence it is preferable to reinforce the top. The lateral 
reinforcement may consist of closely spaced spirals or of bands of 
steel. Some of the patented systems have a special reinforcement 
at the head, as explained in the descriptions of these systems. The 
giant pile is driven by a special method which entirely eliminates 
the stresses during driving. 

The stresses occasioned by uneven driving tend to produce frac- 
ture of the end of the pile. Such stresses take place, for instance, 
when the pile strikes a boulder. The liability to fracture is reduced 
by providing a metal point for the pile and also by using, in addition 
to longitudinal bars, closely spaced lateral reinforcement throughout 
the pile, which reduces the brittleness of the concrete. : 

Concrete Mixture for Pre-cast Pile-——Concrete of 1:2:4 mix 
is ordinarily used for concrete piles. When the pile is driven to 
solid bearing and its capacity depends upon the strength of the pile 
acting as a column, richer concrete may be used with economy. In 
piles relying for their carrying capacity on frictional resistance, the 
use of rich mix would be a waste, as in such cases even the full 
strength of the leaner concrete cannot be fully utilized. 

Details of Design of Pre-cast Piles.— Longitudinal bars should be 
spaced uniformly around the circumference of the pile and should be 
placed 2 in. in the clear from the face of the pile. The percentage of 


CONCRETE PILES 551 


longitudinal steel for column action should be at least 1 per cent of 
the cross section of the pile. Sufficient steel should be provided to 
‘resist bending stresses. In short piles, all the longitudinal rein- 
forcement should extend the whole length of the pile. In long piles, 
where the amount of steel required to resist bending during handling 
is larger than the amount for column action, the bars required as 
column reinforcement should extend the whole length; the balance, 
required for bending only, may consist of short bars stopping where 
the bending moment due to handling permits it. 

Lateral Reinforcement.—Lateral reinforcement may consist of 
separate hoops,. or preferably of a continuous spiral. The spacing 
of the hoops or the pitch of the spiral is made closer at the point and 
at the head than in the central part of the pile. The following 
arrangement is suggested by the authors. In the head of the pile, 
for a distance of 3 ft., provide lateral reinforcement consisting 
of a continuous spiral, the amount of which should form not less 
than 1 per cent of the volume of the enclosed concrete. The pitch 
of the spiral should not be more than 3 in. For the remainder of 
the pile to within 4 ft. from the tip, place separate hoops at least 
¢ in. square, spaced not more than 10 in. apart. At the bottom, in 
the 2-ft. section above the pointed section, provide continuous spiral 
amounting to 13 per cent of the volume of enclosed concrete. The 
same spacing of the spiral should be continued to the end of the pile. 
The total length of the spiral in the pointed part (the toe of the pile) 
and the section above should not be less than 5 ft. 

Pipe for Jetting.—TIf it is intended to use water jet in driving, a 
thin pipe may be imbedded in the center of the pile. Sometimes a 
solid wooden core is well greased and placed in the center of the form. 
This is removed after the concrete has hardened, leaving a hole in the 
center. The core is often tapered, to make the removal easier. It 
it also advisable to turn the core while the concrete is green, so as to 
prevent the concrete from sticking. 

Hangers.—To facilitate handling, provision should be made for 
attaching the tackle. 

‘ Patented Pre-cast Piles.—Several patented Alege are on the 
market. ‘They will be mentioned in the chronological order of their 
appearance. The choice between patented and non-patented pile 
will depend chiefly upon economical considerations of economy. 

Chenoweth Pile is circular in cross section and is built without 
forms. <A 2-in. pipe of a length somewhat greater than that of the 


552 PILES 


pile is placed on a movable table. Wire mesh of proper width 
is attached to the pipe along its length. The mesh is spread 
on the table, and longitudinal reinforcement is attached to it in the 
proper places. A layer of concrete is spread on the mesh. ‘The pipe 
is then revolved, its revolution causing the concrete and the mesh to 
wind around the pipe. The wire mesh in the cross section has the 
appearance of a spiral. 

Gilberth Pile is octagonal in cross section, tapered, and provided 
with semicircular longitudinal grooves in the middle of each face 
and a tapered cored passage in the center. The cored passage serves 
for the introduction of a water jet, and the semicircular grooves are 
intended to afford passages through which the water used in driving 
can come to the surface. After the pile is driven, the grooves are 
intended to increase the friction between the pile and the ground. 

Cummings Pile-—The patented feature of this pile consists of 
a special circumferential and spiral reinforcement at the head, 
intended to prevent splitting or crushing at the top under severe 
impact of the hammer. Tests have proved that the type of rein- 
forcement used serves the purpose. 

Bignell Pile has as its distinguishing feature the use of a double 
jetting system. ‘Two pipes, one within the other, are placed in the 
center of the pile. The larger pipe tapers from 4 in. at the head end 
to about 2 in. at the point. It has no outlet at the end, the water 
being forced to the outside faces of the pile and upward by short 
pipes placed at frequent intervals, at right angles to and connected 
with the main pipe. At their ends, at the faces of the pile, the short 
pipes are provided with right-angle elbows which force the stream 
of water upward. ‘The second pipe, about 2 in. in diameter, is placed 
within the larger pipe. It has a separate inlet at the head and a 
separate outlet at the point. In driving the pile, water is introduced 
into both pipes. High pressure, up to 300 Ib. per sq. in., is main- 
tained in the second pipe, while a pressure of 100 to 150 Ib. is used 
in the large pipe. The water at the end of the pile, coming with 
high pressure, loosens the ground, while the water coming out at the 
sides destroys the friction, so that in some soils the pile sinks under 
its own weight. 

Giant Pile is characterized by a special method of driving, in 
which the concrete pile is not subjected to any driving stresses. 
The concrete pile is square, of uniform cross section, and properly 
reinforced. Its cross section depends upon the load to be carried. 


CONCRETE PILES 993 


Thus far, 16-in. square piles have been most commonly used. The 
concrete pile is firmly attached to a specially constructed cast-steel 
shoe (Fig. 178, below), which at its top projects on all sides be- 
yond the face of the concrete pile. These projections serve as a 
support for a steel driving frame, consisting of two special chan- 
nels, which is provided at the top with a driving head for receiving, 
the blows of the hammer. All the force and the impact of the 
fallng hammer are therefore transmitted by the driving channels 





Fig. 178.—Cast-steel Shoe for Giant Pile. (See p.553.) 


directly to the shoe, permitting very severe driving without injury 
to the concrete. As the shoe progresses, the concrete pile is pulled 
after it. : 

The driving frame is clearly shown in Fig. 179, p. 554. 

In the illustration, the pile is ready for driving, except that the 
driving channels will be lowered until they rest upon the point. The 
length of the driving channels may be greater than that of the pile, 
so that the pile may be driven to any depth below grade, and below 
water. 

When the pile is down, the channels are withdrawn as a unit. 
' The friction upon two sides of the pile at the time of withdrawing 
the channels prevents the upward movement of the pile with the 
channels. The channels are not removed from the pile driver. 

A possible objection to this method of driving is the uncertainty 


554 









































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































PILES 


as to whether, after the withdrawal of the 
frame, the space occupied by it will be filled 
by closing of the soil. This is not of great 
importance when the pile is driven to solid 
bearing. When the pile relies on friction 
for its strength, tests should be made to 
determine whether, in the particular soil, 
the closing is complete enough to offer the 
proper frictional resistance. 

Giant piles have been driven in lengths 
up to 48 ft. in actual construction. In 
experimental work, they have been driven 
up to 60 ft. 

Cast-in-place Piles—As the name im- 
plies, cast-in-place piles are built in the 
position which they are intended to occupy 
in the completed structure. All cast-in- 
place piles are patented. ‘The patents on 
some of them have already expired. The 
most popular of these are described below 
in the order of their appearance. 

Raymond Pile, when completed, consists 
of a sheet-metal shell filled with concrete. 
The shell is tapered. The diameter of the 
shell at the point or bottom is 8 in., and it 
increases at a rate of 0.4in. per foot of 
length of the pile. The diameter of the top 
of the pile, therefore, depends upon its 
length. A 20-ft. pile, for instance, has a 
top diameter of 16 in., while a 30-tt. pile 
has a diameter of 20 in. ‘The shell is fab- 
ricated of black sheet steel, lock seamed, 
spirally corrugated on a 3-in. pitch. In the 
corrugations is placed a 4-in. steel wire, 
which strengthens the shell against collaps- 
ing before concrete is placed and also serves 
as spiral reinforcement for the completed 
pile. The gage of the sheet metal and of 
the wire may be varied to suit conditions. 


Fic. 179.—Giant Pile Ready Ugually, the sheet metal is No. 24 gage and 


to Drive. 


(See p. 553.) 


Ee a a re ee ea 


CONCRETE PILES 500 


the wire ¢ in. diameter. The shell is made up of sections 8 ft. long, 
lapping each other. Raymond pile is shown in Fig. 180, p. 556. 

Ordinarily, the concrete within the concrete pile is not pro- 
vided with vertical reinforcing bars, as they are not necessary to 
resist the compression stresses. When, however, there is any 
reason to expect that the pile will be subjected to bending due to 
lateral forces, reinforcing bars may be placed in the shell prior to the 
pouring of the concrete. 

The pile is constructed in the following manner: The shell is 
assembled to give the proper length. A collapsible steel mandrel 
is inserted into the shell and expanded to fit it tightly. The shell, 
with the mandrel, is then driven by means of a pile driver until the 
desired minimum penetration is attained. The mandrel is then 
collapsed and pulled out, and the shell is filled with concrete. Prior 
to the pouring of the concrete, the shell may be inspected. The 
decided advantage of the Raymond Pile over other cast-in-place piles 
is that the shell prevents water and earth from mixing with the 
concrete. It also protects the green concrete from damage from 
lateral pressure of the earth. The shape of the pile is, therefore, 
certain. or piles relying for their capacity on bearing, the taper 
of the pile is a disadvantage, as the cross section at the point is 
small. Raymond Piles have been built in lengths as great as 37 ft. 
6 in. 

Raymond Composite Wood and Concrete Pile consists of a wood pile, 
driven below permanent low water, upon which is superimposed a 
Raymond Concrete Pile. Before the wood pile is driven, the top of 
it is provided with a tenon 95 in. in diameter and 18 in. long, squarely 
cut on the top and fitting closely into the opening in a specially pre- 
pared steel shell. A reinforcing bar, attached to the tenon, extends 
into the concrete pile and joins the two parts. Fig. 181 shows the 
junction of the wood pile and the concrete pile. 

The wocd pile is driven to ground level. The collapsible man- 
drel, encased in a spirally reinforced steel shell, is then fitted on the 
top of the wood pile. The combined pile is driven to its final pene- 
tration. The mandrel is withdrawn and the shell filled with con- 
_ crete. : 

Simplex Pile, in its finished state, consists of a plain concrete 
shaft of a constant cross section throughout, usually about 16 in. in 
diameter. It is provided with a cast-steel or cast-concrete point. 
(See Fig. 182, p. 557.) The pile is built by fitting the cast point into 


PILES 


596 


(‘egg daa): 
‘OTT POOM aytsodur0y) puowAvy— TST “OM 





(‘Fag “d 299) 


‘gTq puow Avy 











CONCRETE PILES 557 


the bottom of a steel tube of the desired diameter, and driving them 
in the required position until the minimum penetration is attained. 
The steel tube is then gradually withdrawn by means of -a pulling 
tackle, and at the same time the space is filled with concrete. Sim- 
plex piles have been used up to 50 ft. in length. 

Particular care should be taken in filling the pile, not only to_ 
avoid arching of ‘the concrete, but also to prevent the withdrawing 
of some of the concrete with the tube, 
thus leaving a void in the pile, which 
would destroy its value. 

Pedestal Pile has the theoretical shape 
shown in Fig. 183, p. 558. It consists of i" 


, Pulling ee 


y Pulling Clamp 


SSSSSsssssssss 


a 


-~- Steel Driving 
Form 


CSSSSpasssssssss 


a plain concrete shaft and a bulb-like 
enlargement or pedestal, which increases 
the bearing area of the pile. 

It is built in the following manner: A 
steel casing, usually 16 in. in diameter, 
with a solid plunger fitted into it, is 
driven by a pile driver to the desired 
depth. The plunger is withdrawn and 
the casing partly filled with concrete. 
The plunger and a hammer are then 
dropped into the casing, forcing the 
concrete sidewise against the soil and 
gradually enlarging the bottom space. 
Successive fillings and rammings complete 
the bulb. After this, the rest of the pile 
is filled, and at the same time the steel 





Concrete 


MINS: é ere : 
casing is gradually withdrawn. ‘To pre sea rie 
vent the upward movement of the pile Pulled 

with the casing, the plunger and the Fic. 182.—Details of Simplex 


hammer rest on the concrete while the Pile. (See p. 555.) 
casing is withdrawn. 

T'rom the method of producing the bulb, it is evident that its 
shape depends altogether upon the compressibility of the soil. It 
will be of uniform shape if the soil is homogeneous; otherwise, the 
soil may compress unevenly and result in an unsymmetrical bulb. 
This may produce eccentric stresses in concrete 

Peerless Pile, when in place, consists of a cast-steel shoe and a 
thin concrete shell of uniform cross section resting on the shoe and 


558 PILES 


solidly filled with concrete. The shell is made up of sections and 
extends the full length of the pile. 

The pile is built in the followmg manner: The point, with the 
superimposed shell, is driven by means of a plunger to the required 
penetration. Then the plunger is withdrawn and the shell is filled 
with concrete. To facilitate driving, the area of the top of the shoe 
‘5 somewhat larger than the area of the shell. The hole made by the 
shoe is therefore somewhat larger than the shell. This makes it 
unnecessary to exert any pressure on the shell in order to cause it to 
follow the shce. : 


Head of Core 
( _-- Casing 
/(Shown in Cross - Section) 










Surface of Ground 














Fic. 183.—Details of Pedestal Pile. (See p. Note 


This pile has the same advantages as the Simplex Pile. The 
shell protects the green concrete from lateral pressure. 

Miscellaneous Piles —Other piles have been developed, but their 
limited use does not warrant their description. 

Carrying Capacity of ‘Cast-in-Place Piles—The  cast-in-place 
piles rely, for their carrying capacity, either on the bearing of their 
points, or on friction. Ina majority of cases, the concrete piles are 
designed for 30 tons. 

- Piles that rely on the bearing of their points may be considered 
as columns, and their strength should be determined in the manner 
explained in connection with pre-cast and wood piles. Under such 
conditions, the piles without taper are stronger than the tapered 
piles. 


CONCRETE PILES 559 


The capacity of cast-in-place piles that rely for their bearing 
power on friction may be determined by means of the Engineering g 
News Formula, p. 544, with the same accuracy as for wooden piles. 
The penetration, the weight. of hammer, and the fall are those used in 
driving the core. Some building codes require load tests on concrete 
piles not driven to rock. 

Advantages and Disadvantages of Cast-in-Place Piles—The cast-in- 
place piles can be built’ without delaying the construction. They 
do not require any storage space, nor expensive handling equipment: 
hence, they are particularly adapted for use where piles must be 
driven in built-up localities. As the length of the piles can be well 
regulated, it is not necessary to predetermine it by borings, as in the 
case of pre-cast piles. 

The disadvantage of cast-in-place piles is that their shape cannot 
be inspected after erection. This applies particularly to piles built 
without a shell. Another disadvantage is that, unless the progress 
of driving is properly regulated, the setting of the concrete may be 
affected by the vibration caused by driving adjacent piles. The 
effect of vibration is harmful from the time of initial set, i.e., from 
about three to four hours after pouring until the concrete is thor- 
oughly hard. 

Piles for which the shell remains in the ground are superior to 
other cast-in-place piles. If the shell is strong enough, the green 
concrete is protected from injury by lateral pressure of the earth 
produced by the driving of adjacent piles. No foreign matter can 
enter the shell. When filled with proper care, the completed pile is 
of the shape contemplated. 

The use of piles from which the casing is removed should be 
restricted to stiff soils, capable of retaining the shape of the hole 
until the concrete hardens. In numerous instances, excavation of 
piles has shown that their shape was distorted to such an extent as to 
render them useless.! Piles of this type should not be used where 
the chemical composition of the soil is such as to affect the setting 
qualities of the cement. Neither should they be used where the 
ground is likely to flow, nor where water would introduce admix- 
tures of the earth into the concrete. In passing water-bearing 
strata, in piles without shell, the cement is likely to be washed out. 

1 See History of the Concrete Pile Industry, by Charles R. Gow. Proceedings 


Am. Concrete Institute, Vol. XIII, 1917, p. 202. Also Engineering Record, 
March 22, 1913. 


560 PILES 


In soils transmitting lateral pressure, the green concrete may be 
distorted by the pressures due to the force used in driving adjacent 
piles. . 

In all cast-in-place piles, special care must be used to prevent 
the formation of voids in the pile. Where the casing is gradually 
removed, care should be taken to prevent lifting of the concrete pre- 
viously placed with the casing, as this would result in reduction of the 
cross section or separation of the pile into sections. | 

Caisson Pile.—The caisson pile usually replaces a whole cluster 
of piles. One pile is ordinarily used per footing. A caisson pile,” as 
shown in Fig. 184, p. 560, con- 
sists of a cylindrical shaft of 
plain concrete, strong enough 
to carry the column load, 
reaching from the bottom of 
the basement or ground floor 
slab to the bearing soil, where 
it is enlarged to provide the 
required bearing area on the 
soil. This enlargement, often 
called the bell, is in the shape 
Section A-A of the frustum of a cone. The 

slope of the sides, usually 60° 

with the horizontal, is such 

that the stresses in the pro- 

jection due to the soil pressure 

= do not exceed the value permit- 
* 

~~~ Supporting Soil ted for plain concrete. The 

Hie, 184.-—Caisson Pile. (See p. 560.) allowable inclination for differ- 

ent pressures may be taken 

from the table on p. 481. If desired for any reason, the pier may 

be provided with vertical reinforcement. Horizontal reinforcement 

may also be used at the bottom of the bell if required by the stresses. 

At the lot line, the enlargement of the pile may extend only so 
far as to be parallel to the lot line. To prevent eccentricity, the 
wall column may be connected with the first row of interior columns, 
by means of a strap beam. 

The excavation for a caisson pile is made by sinking into the 







--= Column 





2In the eastern part of the United States, they are often called Gow piles, 
from Charles R. Gow, who introduced them there. 


CONCRETE PILES 561 


ground short cylindrical sections of proper diameter, and simulta- 
neously excavating the material within with pick and shovel. After 
the-material is excavated to the full depth of the cylindrical section, 
another cylindrical section is driven, and this is repeated until the 
desired level is reached. The excavation is then enlarged to form the 
bell. No form is used for the sides of the bell. After the excavation 
is completed, concrete is deposited. It is desirable to pour the con- 
crete into the whole pile, or at least into the bell-shaped portion, by 
a continuous process. As the depositing of the concrete progresses, 
the cylinder sides are removed. 

The hollow cylinder may be made of wood staves or of sheet 
metal of proper thickness. The sheet metal may be in one piece or 
in several vertical sections bolted together. The wood forms are 
held in position by collapsible steel rings placed inside. 

The smallest diameter of the shaft that is convenient for a man 
to work is considered to be 3 ft., which is strong enough to carry 
508 000 lb. with a unit stress of 500 Ib. per sq. in. Shafts for columns 
carrying smaller loads will have the same diameter. It is important 
to remember, however, that even if the same size of shaft is used 
for different loads, the area of the bell must be proportioned accord- 
ing to the load to be carried. 

Thus, if the column loads (including weight of pile) are 400 000 lb., 
450 000 Ib., and 500 000 Ib. respectively, the required diameters of 
the bell for the different columns, with an allowable soil pressure of 
8 000 Ib. per sq. ft., will be 8 ft., 8 ft. 6 in. and 9 ft. respectively, 
even if the shaft in all cases is 3 ft. in diameter. 

Although the caisson pile is built in place, it is free from the dis- 
advantages of the built-in-place single pile. The excavation can be 
thoroughly inspected before concreting, and the forms are sure to 
stay in place while the pile is being filled. As no pile driver is used 
in driving adjacent piles, no lateral stresses are set up in the green 
concrete. Neither is the concrete injured in setting by vibration. 


Example 3.—Design a caisson pile to support a column load of 800 000 Ib., 
using maximum allowable stress on concrete of 500 lb. per sq. in. and maximum 
bearing on soil of 8 000 lb. per sq. ft. 


Solution.—The required area of the pier, to carry the load, equals 800 000 + 
500 = 1600 sq. in. To give this area, a 45-in. round pier will be selected. 

Assuming dead load of the pile as 60 000 lb., the total load on soil is 860 000 Ib. 
The area required for bearing is: 

860 000 + 8 000 = 108 sq. ft. The required diameter of the bell is 12 ft. 3 in. 


562 PILES 


The inclination of the sides of the cone may be found as explained on p. 481. 
For a unit pressure of 8 000 Ib., the height of the cone equals 1.7 times the 
projection of the bell. 


Steel Cylinder Foundation.—Unlike the cast-in-place concrete 
piles, which rely wholly on concrete for strength, the steel cylinder 
foundation, such as the “ Hercules ond “ Steel Tuba ” foundations, 
derives a large part of its strength from steel cylinders driven to solid 
rock. The diameter of the cylinders is 10, 12, 15, and 16 in., and the 
thickness of the cylinder wall not less than 3 in. 

The cylinders are driven to solid rock by means of a steam 
hammer. Where the distance to rock exceeds 22 ft., the cylinder is 
driven in 20-ft. sections, and the sections, as driven, are joined by 
specially designed water-tight sleeves. The material is removed 
from the cylinder by compressed air. After the cylinder reaches 
solid bearing in rock, it is cleaned and filled with concrete. It is then 
cut to grade and leveled by an oxy-acetylene torch. An I-beam cap, 
of the dimensions given below, is placed on top to transfer the pres- 
sure from the footing to the cylinder, A reinforced concrete footing 
is built on top of the cylinder. 


Dimensions of I-beam Caps 


Diameter of Cylinder Depth of I-beam Length of I-beam 


16 in. : 20 in. 20 in. 
15 in. 18 in. 18 in. 
12 in. 15 in. 15 in. 
10 in. 15 in. 15 in. 


When driven to rock, the capacity of a steel cylinder pile is 
determined by its safe strength. The New York Building Depart- 
ment permits the following loads on steel cylinders: 


Capacity of Steel Cylinder in Tons 


ee 








Outside Diameter of Cylinder 








Thickness 
of Steel, In. ane | 
103 in. 123 in. 15 in. 16 in. 
es ES 
a 56.7 73.6 93.5 102.9 
<5 65.4 82.9 103.2 113.4 
4 73.4 92.3 112.8 123.27 


CONCRETE PILES 063 


In computing the strength of the cylinder, the assumption is 
made that rust will penetrate for #3 in. Therefore, the effective 
thickness is 7's in. smaller than the actual thickness of the cylinder. 

Steel cylinders can be used only where they can be driven to solid 
rock. They are particularly advantageous in foundations built 
close to adjacent structures, where owing to the compactness of the 
foundation, it is often possible to get along without combined foot- 
ings. 


CHAPTER X 
BUILDING CONSTRUCTION 


Reinforced concrete has taken its place as an established material 
for building construction. Durable, fireproof, and economical in 
first cost, adaptable to various types of design, capable of carrying 
heavy loads, and at the same time susceptible of pleasing architec- 
tural treatment, it holds a unique position as a building material. 

For the structural frame of factory and office buildings, for foun- 
dations and floors of steel frame structures, or as artificial stone for 
facing or trimming, its adaptability is recognized. For small build- 
ings such as dwellings, its use is not so general, because of larger unit 
costs on small jobs; but in certain cases where, on the one hand, 
expense is not the criterion, and, on the other hand, where duplica- 
tion of design reduces the cost, it is being adopted to advantage. 

For first-class construction there are three requisites: (1) thor- 
oughly tested materials; (2) design by an engineer familiar with 
reinforced concrete design; and (3) construction by an experienced 
builder working under careful supervision. 


RELATIVE COSTS OF BUILDINGS OF DIFFERENT MATERIALS 


For industrial and office buildings, reinforced concrete naturally 
competes with the steel frame, plain or fireproofed, and with mill 
construction. Cost is usually the important factor; but sometimes 
speed after breaking ground is the main consideration, as is the case, 
for example, in high buildings in the business sections of large cities, 
and structural steel may be selected on this account. If the time of 
fabrication of the structural steel must be included, however, the 
concrete building can be put up in a shorter time. - 

In selecting the type of building, it is necessary to consider not 
only the first cost, but also the average annual expense and deprecia- 

564 


RELATIVE COSTS OF BUILDINGS OF DIFFERENT MATERIALS 565 


tion over a term of years. It may be economical to increase the first 
cost for the sake of an annual Saving in expense; if this saving will 
be sufficient, in the course of the useful life of the building, to make 
up for the higher initial expenditure. | 

Fireproofed steel frame construction is almost invariably more 
expensive in first cost than reinforced concrete. 

The first cost of the reinforced concrete structure, in turn, may be 
greater than that of a steel frame, not fireproofed, or of mill con- 
struction. This depends, however, to a considerable extent, on the 
type of building. Thus, with very heavy loads, especially on long 
Spans, concrete is cheaper than steel or mill construction. The 
dividing line varies with relative costs of material. Frequently it 
occurs at loads of 200 Ib. per sq. ft. on spans in the neighborhood of 
20 ft.! 

To get any real comparison between buildings of different 
materials, a detailed estimate should be made of the first cost and 
the annual expenditures. The following table,? issued by the Uni- 
versal Portland Cement Company,? illustrates the method: 


Comparative First and Maintenance Cost of Reinforced Concrete and Mill 
Constructed Buildings 


(From standpoint of Owner) 








Reinforced Mill 
Concrete Construction 
(Fireproof) (Not Fireproof) 
First cost of building............ $189 000.00 $168 000.00 
First cost of sprinkler system.... . 14 000.00 14 000.00 
Total-first investment.......... $203 000.00 $182 000.00 
First cost fireproof more than mill construction................. $21 000.00 


‘See paper by L. C. Wason in Proceedings National Association of Cement 
Users, Vol. VII, 1911, p. 448. 
| * Similar computations are given by J. P. H. Perry in Proceedings National 
Association of Cement Users, Vol. VII, 1911, p. 443. 
* See also Concrete Cement Age, July, 1916, p. 25. 


566 BUILDING CONSTRUCTION 


Maintenance 

Interest on first investment... .6[% $12 180.00 $10 920.00 6% 
Tax on first investment.....-- 1% 2 030.00 1 820.00 1% 
Depreciation on building. . . 0.5% 945.00 3 360.00 2% 
Obsolescence. 00.5 63.4 < 5 6 aie 1.0% 1 890.00 3 360.00 2% 
Depreciation on sprinkler .....10% 1 400.00 1 400.00 10% 
Repairs to building........ 0.25% 472.50 1 680.00 1% 
Damage to building by vermin 

Nond.t. ont ee eee ee 200.00 Est. low 
Auxiliary fire equipment. 

Estimated gag sadudiena> tee See 200.00 300.00 Estimated 
Fire insurance on building. 

None required. .....---+---+++: 235.20 14 cts. on 

$100 
$19 117.50 $23 275.20 
Yearly expense fireproof less than non-fireproofi. es Aare $4 157.70 


The yearly saving of $4 157.70 capitalized at 6 per cent represents $69 295. 
Therefore, actual cost of concrete building is $119 705, in comparison with one of 
mill construction costing $168 000. 

Furthermore, there is a lower rate for fire insurance on the contents of the 
concrete building, which still further reduces the cost. 


Reinforced concrete has been used economically for dwelling 
houses, but only where cheap cottages can be built in groups of similar 
pattern. With this exception, wood is cheaper, on account of the high 
cost of forms. Precast blocks, requiring no forms, are simplest for 
this class of work, but unless the surfaces are tooled the appearance 
is apt to be monotonous. In estimating the labor where forms are 
used, allowance must be made for time lost in waiting for the con- 
crete to harden so that the forms can be raised. For this reason, a 
small gang of men should be employed—only enough to lay concrete 
to the height of one section of the forms per day. 

For cellar and foundation walls of all classes of buildings, including 
brick and frame, concrete is superseding rubble masonry except 
where rubble stone is taken from the excavation so as to be very 
cheap. 

Cement mortar plastered on to metal or wood lathing is used not 
only for outside walls, but in some cases for fire-resisting partitions in 
large buildings. 





ACTUAL COST OF REINFORCED CONCRETE BUILDINGS 567 


ACTUAL COST OF REINFORCED CONCRETE BUILDINGS 


The tables presented on page 568 give the approximate average 
cost per square foot of floor area and per cubic foot of volume of 
plain rectangular reinforced concrete buildings of various sizes and 
heights. 

The costs include all details of construction, not only the concrete 
forms and reinforcement, but also windows, stairs, roof covering, and 
plumbing. Interior finish, which varies widely with the type of 
construction, is not included. The basis * of the tables is as follows: 

(1) Floor loads, 150 pounds per square foot. 

(2) Story heights: first floor on a 3-foot fill; 

other floors 12 feet from slab surface to slab surface. 

(3) Column spacing, 18 feet on centers. 

(4) Floor design: girders between columns in one direction; beams between 

columns in other direction with two intermediate beams. 

(5) Excavation and foundations. + 

















Story Height — 


Outside Walls, Inside Walls 
per Linear Foot | per Linear Foot 





$5.45 45 $2.80 
6.85 5.00 
7.20 6.00 
7.60 7.05 
7.90 8.45 


| 
Sha 
6.15 3.75 





(6) Filling under first floor: 3-foot fill at $1.00 per cubic yard in place. 
(7). Stairs: material and labor, $200 per flight per story. 
(8) Stairways and elevator towers: 
2 stairways and 1 elevator tower for buildings up to 150 feet long. 
2 stairways and 2 elevator towers for buildings up to 300 feet long. 
3 stairways and 3 elevator towers for buildings over 300 feet long. 
(9) Floor finish: all floors of concrete with granolithic finish. 
(10) Walls: 
(a) Curtain walls between pilasters, 3 feet high and 8 inches thick; 
(b) Concrete walls for penthouses, 6 inches thick. Dimensions of pent- 
house are 10 feet by 10 feet; 
(c) Concrete walls around the elevator and stairway openings are taken 
6 inches thick, the elevator opening being 10 by 20 feet and the 
* Values fixed with the advice of Morton C. Tuttle Co. They conform to 1925 prices. 
{ Taken from paper presented before the New England Cotton Manufacturers’ Association, 


April, 1904, by Mr. Charles T. Main. Prices revised by Mr. Main to conform to prices prevail- 
ing about January, 1925. 


568 BUILDING CONSTRUCTION 


Average Costs of Concrete Buildings per Square Foot of Floor Area (See p. 567) 
Costs include all items except interior finish ) 


ea a eee ae eee ae 





Cost in Dollars per Square Foot of Floor Area 








ee ee 


Width 
Feet Length of building, in feet Length of building, in feet 
50 | 100 | 200 | 300 | 400 | 600 || 50 | 100 | 200 | 300 | 400 | 600 
1-Story 2-Story 
25 4.94; 3.86) 3.38 3. 08! 9.95) 2.91|| 4.83, 3.73; 3.27, 3.02 2.89! 2.74 
50 | 3.52 3.02) 2.66] 2.40! 2.28/ 2.21 3.46 2.74| 2.43| 2.21| 2.13) 2.07 
poe 8 91| 2.78) 2.43) 2.17 2.07| 2.00|| 3.04) 2.51) 2.17 2.02 1.921 1.83 
100 3.04| 2.62]. 2.28] 2.07 1.92) 1.88|| 2.85 2.32) 2.04) 1.88 1-77 P71 
150 2.93| 2.49} 2.17| 1.96 138)..1 77 2. 68| 2.19) 1.92) 1.75) 1.67 1.60 
4-Story 6 to 10-Story 
25 4.68| 3.54| 3.08| 2.89] 2.76| 2.64|| 4.68) 3.50| 3.06 2.85! 2.78} 2.64 
50 3.25| 2.53] 2.26) 2.11) 2.05 1.96|| 3.23] 2.49| 2.24) 2.11] 2.05) 1.96 
75 2.85) 2.28} 2.02) 1.90 1.83) 1,77} 2281) 2228) 2.02 1.88] 1.79] 1.75 
100 2.64| 2.13} 1.88] 1.75 1.69| 1.64]| 2.62} 2.09) 1.86 1.73| .1.67| 1.62 
150 2.49} 2.00] 1.77| 1.64 1.88] 1.52|| 2.45) 1.96) ders 1.62| 1.58) 1.52 


First floor on fill. Slab supported on the ground. 
To use the table multiply the aggregate floor area (exclusive of the roof area) by the unit 
costs. 


Average Costs of Concrete Buildings per Cubic Foot of Volume (See p. 567) 
Costs include all items except interior finish 


Cost in Dollars per Cubic Foot of Volume 








Width 
Fee t Length of building, in feet | Length of building, in feet 
Ee ee eee 
50 | 100 | 200 | 300 | 400 | 600 50 | 100 | 200 | 300 | 400 | 600 
1-Story 2-Story 


pee ee ee eee 
25 0.41110. 323/0.281|0.257/0.247/0. 243 0.403|/0.310 
50 0. 293/0.251/0. 2210. 200)0.190)0. 183 0.289)0.228 


0.167||0.253|0.209/0.183)0. 169)0.160\0. 152 
100 0. 253/0.219/0.190/0. 173/0. 160)0. 156 0. 238/0.194/0.171|0. 156/0. 148)/0. 141 
150 0. 245/0.207|0. 181/0. 162/0.152/0.148 0. 224/0.183/0.160|0. 145|0. 139]0. 133 


4-Story 6 to 10-Story 


25 10.390|0.295/0. 257/0.240/0. 230 0.219)|0.390/0. 291/0. 255)0. 236)0. 232|0.219 
50 0.270|0.211)0.188|0.175|0.171)0. 162 0.270\0.207|0.186)0.175|0.171/0.162 
| 
0.190|0.169|0.156|0.150|0.145 
0.173|0.154/0. 143]0.139]0.135 
0.162\0.143\0. 1385/0. 13110. 127 


0.272)0.251|0.240'0.228 
0.202|0.186/0.177|0.173 





75 0.26610. 232\0.202/0.181)0.173 











75 0. 236/0.190,0. 169]0. 158)0.152/0.148 0.234 
100 0. 21910.177|0. 156|0. 145\0. 141)0. 137 0:217 
150 0.20710. 167|0. 148/0. 137/0.133|0. 127 0.205 























Values are based on conditions outlined on page 567. The tables are taken from ‘‘ Concrete 
Costs”” by the same authors, converted to 1925 prices, and the values are made up from 
tables of unit times and costs given in the same book carefully checked by contractors’ esti- 
mates. For more complete details and for the unit values which are adapted to all conditions, 
see other tables and examples in “‘Concrete Costs.” 

Values are for symmetrical buildings. 


FLOOR LOADS 569 


stairways 10 by 10 feet, these two being adjacent so that the one 
intermediate 10-foot wall serves for both openings; 

(d) For toilets, concrete walls 6 inches thick and 20 feet long, one wall for 
each 5 000 square feet of floor space. 

Walls 8 inches thick, including reinforcement and forms, 75¢ per square 


foot. 
Walls 6 inches thick, including reinforcement and forms, 68¢ per square 
foot. 
(11) Windows and doors: all openings for windows and doors, 75¢ per square 
foot. 


(12) Roof covering and flashing: five-ply tar and gravel roofing, 60¢ per 
square foot. . 

(13) Plumbing: two fixtures on each floor up to 5000 square feet of floor 
surface, and one additional fixture for each additional 5000 square 
feet, $150 per fixture. 

(14) Labor rates: carpenter labor, $1.10 per hour; steel labor, $1.10 per 
hour; and common labor, 65¢ per hour. 

(15) Concrete in place (including labor and materials): $12.00 per cubic yard, 
or 44¢ per cubic foot. 

(16) Form lumber: $45.00 per 1 000 feet B. M., delivered. 

(17) Steel for reinforcement: $55.00 per ton, delivered. 


For lighter loads than specified, the costs are slightly decreased, 
this decrease running up to 25 cents per square foot for a 75-pound 
load in a 10-story building. For a 300-pound load, the prices are 
increased from 12 cents for a 2-story building up to 25 cents per 
square foot for 10 stories. It must be remembered that the tables 
are based on rectangular, symmetrical buildings. Allowance must 
be made for irregular layouts, which increase materially the cost of 
form construction. , 

The variation in cost due to variation in spacing of columns is 
small. If columns are spaced 15 feet apart the cost is 6 per cent 
greater than where columns are spaced 25 feet apart. 


FLOOR LOADS 


Live Loads.—In designing any structure, it is necessary to deter- 
mine the design live load. This should be estimated from the investi- 
gation of the use for which the building is designed. The live load 
must not be smaller, however, than required by the building code of 
the city in which the building is erected. It should be borne in 
mind that the codes usually give the minimum values. If, from 
investigation, it should follow that the expected live load in the build- 
ing under consideration is larger than the live load required by the 
code, the larger live load should be used. 


070 BUILDING CONSTRUCTION 


To illustrate good practice, the following requirements of the 
Boston Building Code are given: 
Live Load Required by Boston Building Code. 







Pounds per 


Class of Buildings Square Foot 








a ae ee ee Ts ts 


Armories, assembly halls, and gymnaSiUMS......---.eeeee eee eee 100 
Fire houses: 
Apparatus floors.......--- Save, {ey hs a Se 150 
Residence and stable floors. .....2.)5-- 5. 00 eee ee 50 
Garages, private, not more than two ¢ars. ei Os a ee 75 
Garages, public.) 0.0 0. PS 150 
Grandstands ic. o. $0. oye be 6 i oo 100 
Hotels, lodging houses, boarding houses, clubs, convents, hospitals, 
asylums and detention buildings: 
Public portions... 0h SEE a ee 100 
Residence portions....... .¢.0. 2047 c <4 0 a> 7s se PDS 
Manufacturing, heavy s.. 20 se. oe oe Ne eens Fim se 250 
Manufacturing, light. 2M O eee ee 125 
Office buildings: 
First floor...sy.eecueeeeser sete nee nn 40s seco ee 125 
‘All other floors..«. iv.n0+ 2 s* 2h sree Been ates oe Vc Wud 75 
Public buildings: 
Public portions). £105. oy 1200s. anaes oh «ate alec 100 
Office portions: ...ag%s fneeiene oett ba aus FR. + Rs See stem 75 
Residence buildings, including porches......-.++-+++++esrtees07? 50 
Schools and colleges: 
Assembly halls... ¢22¢---4.- 0+ sess cee a) rr 100 
Class rooms never to be used as assembly halls... 7720"). eee 50 
Sidewalks: 
(or 8 000 pounds concentrated, whichever gives the larger moment 
or shear) .....-..+o+ 4209 es bimpee ns tote adel Rei 250 
Stables, public or mercantile: 
Street entrance floors ......-+4te see 74-550 ei 150 — 
Feed roonit...,.1.c.00+seees seen eres et are +o oe tei 150 
Carriage roomie i. ++ +l. 0s 5 Satna ie 2S 50 
Stall TOO ..ntasct 4 oa byacpehee eee eee Perper 8 50 
Stairs, corridors, and fire escapes from armories, assembly halls, and 
pymNasiumng.\ss..cres ++ ts feee ee ene: es ee aperae 100 
Stairs, corridors, and fire escapes except from armories, assembly 
halls, and gyrimasiums. . ./))\75 .s yo) sigh se eee 75 
Storage, heavy. slci+seaest osc stk: ae rh: oe eee te 250 
Storage, light... 0.4). espa phe Sab or 125 
Stores, retail... ce y oss <a i sph: algae SS: saree 125 
Stores; wholesale. .00 05... ere us tents a os oe 250 





FLOOR LOADS O71 


The Boston Code provides further as follows: 

Every plank, slab, and arch, and every floor beam carrying one hun- 
dred square feet of floor or less, shall be of sufficient strength to bear 
safely the combined dead and live load supported by it, but the floor 
live loads may be reduced for other parts of the structure as follows: 

In all buildings except armories, garages, gymnasiums, storage 
buildings, wholesale stores, and assembly halls, for all flat slabs of 
over one hundred square feet area reinforced in two or more direc- 
tions and for all floor beams, girders, or trusses carrying over one 
hundred square feet of floor, ten per cent reduction. 

For the same, but carrying over two hundred square feet of floor, 
fifteen per cent reduction. 
| For the same, but carrying over three hundred square feet of 

floor, twenty-five per cent reduction. 

These reductions shall not be made if the member carries more 
than one floor and therefore has its live load reduced according to the 
table below. 

In public garages, for all flat slabs of over three hundred square 
feet area reinforced in more than one direction, and for all. floor 
beams, girders, and trusses carrying over three hundred square feet 
of floor, and for all columns, walls, piers, and foundations, twenty- 
five per cent reduction. 


Reduction of Live Load in Column Design.—The rules for reduc- 
tion of live load used for design of columns is given in the chapter on 
Columns, p. 453. 

Dead Loads.— Dead loads consist of the weight of the construction 
and of all fixed loads, such as floor finish, plaster, partitions, walls, ete. 

The dead loads may be taken from foilowing table: 


Weight, 
Description Pounds per 
Square Foot 
Granolithic finish per inch of thickness (see p. 620).............. 12 
Hardwood floors: 
% hardwood top, +2 intermediate floor, screeds, 2 in. cinder con- 
crete fill (see p. 625) sg Si Paap De ne SER eer ERO Mt | 251 
Same except 1} intermediate planking (see p. 625)............. 23 
Same except intermediate floor omitted (see p. 625)............ 18 
% hardwood top, intermediate floor, 13 plank and tar base (see 
TERED Go SE INS lh a eas a See lee ae 5 ie ee Maa 16 
3-in. wood block floor in coal tar pitch (see p. 626)............... 10 
eee Ate GEIGK 0... open: Aexecws opt oh.+,+ #0, <cugeiata Rn a 14 
Penamrer Cueconerete OF 1116 {2°COALS) 0 oe eu = ge ee ce ye ee eee 5 
PUM MIRCH ORC a Mr, Loe. Lo aty ali'e cas Wn cn One ee Pe 10 


572 BUILDING CONSTRUCTION 


GENERAL DESCRIPTION 


Reinforced concrete buildings may be either of the skeleton or 
the wall bearing type. 

































































Fic. 185.—Skeleton of Salada Tea Building. (See p. 573.) 


In the skeleton type, the reinforced concrete members form a 
self-sustaining skeleton consisting of concrete foundation or footings, 
columns resting thereon, and floor construction supported by the 


columns. The enclosing walls, or “curtain walls,” are supported 


GENERAL DESCRIPTION 573 


by the skeleton, and may be built after the skeleton is completed. 

In the wall-bearing type, the walls support the floor construction. 
The walls may be of brick, concrete, concrete block, or terra cotta 
blocks. The description of this type is given on p. 629. 





Fig. 186.—Perspective View of Beam and Girder Skeleton. (See p. 574.) 


Skeleton Type.—A photograph of a completed skeleton of a re- 
inforced concrete building is shown in Fig. 185, p. 572. The compo- 
nent parts of the building are plainly seen. ‘The floor is of the flat slab 
design. Fig. 263, p. 752, shows the same building after it is enclosed 
by walls. The concrete skeleton is completely covered. In less orna- 


574 BUILDING CONSTRUCTION 


mental buildings, the concrete columns or concrete spandrel beams, 
or both are exposed. (See p. 729.) 

A clear understanding of the construction of a skeleton may be 
had from Figs. 186 and 187, showing, in perspective, sections of two 


4 





oe Wall Bearn 


Sas 










\ “Drop Panel 
“Column Head 


























--- Column 





dlab on Fin 


Fiq. 187.—Perspective View of Flat Slab Skeleton. (See p. 574.) 


buildings. In Fig. 186 the floor is of beam and girder design, and in 
Fig. 187 of flat slab design. The names of the component. parts, 
which are described under proper headings in this section, are marked 

- on the figures. | 
Types of Floor System.— Floor construction is of two general 
types: the beam and girder design, and the girderless or flat slab 


BEAM AND GIRDER DESIGN OF FLOOR O75 


design. Hach of these may be further subdivided as indicated in the 
detail treatment. 


BEAM AND GIRDER DESIGN OF FLOOR 


The beam and girder design is the oldest type of concrete floor 
construction. It follows the scheme used in steel construction, the 
load being carried by the slabs, which are supported by beams. The 
beams either run into columns or are supported by girders running 
into columns. The arrangement of beams and girders most usual in 
practice is shown in Fig. 188. 

The light-weight floor systems described on p. 588 are modifica- 
tions of the beam and girder design. | 

Use of Beam and Girder Design.—Originally, the beam and 
girder floor construction was used almost exclusively. Now its use is 
restricted by the development of the flat slab floor and the light- 
weight floor systems. The beam and girder floor is used for heavy 
loads in preference to flat slab construction, under the following cir- 
cumstances: (1) when the enlarged heads of the flat slab type are 
objectionable, irrespective of the dimensions and arrangement of 
panels; (2) for oblong panels where flat slab construction is not 
feasible; (3) for very long spans; (3) when the structure is only one 
panel wide; (5) when the floor has many openings. For light loads, 
it is used in preference to light-weight floor: when the tile fillers are 
not readily obtainable; when there is a possibility of concentrated 
loads; when the appearance of exposed Joists is objectionable and 
the use of furred ceiling is too expensive; when the use of furred 
ceiling is not warranted and the building is exposed to large fire 
hazards. 

Beam and girder construction is also required in flat slab floors 
around staircases and elevator openings, and in odd panels. 

Economical Arrangement of Beams.—The economy of construc- 
tion depends to a great extent upon the arrangement of beams in a 
panel. In determining the most economical arrangement, not only 
the cost of materials, but also the cost of formwork must be consid- 
ered. Close spacing of beams permits the use of lighter slab, thereby 
reducing the amount of concrete and steel, but it increases the cost of 
formwork. Wide spacing of beams saves formwork, but. increases 
the amount of materials. An economical arrangement is that in 
which the sum of the cost of materials and the cost of formwork is a 


BUILDING CONSTRCUTION 


576 


f the cost of labor 


Where materials are scarce, but labor is 


It will vary with changing ratio o 


minimum. 


to the cost of materials. 





EE 
4-p----- Bac 


| 


Jj-L--------' 
1 








oe Se 








1 
ae a i 


-~-=--——----------4 


| 

! 

! 

| 

J —_— 
r 
rc 
| 

' 

' 

| 





Fic. 188.—Arrangement of Beams and Girders. 


(Sze p. 578.) 


it is more economical to save materials at the expense of extra 


formwork. The reverse is a 


cheap, 


Iso true. 


d also upon the relation between 


Economical design will depen 


BEAM AND GIRDER DESIGN OF FLOOR O77 


the prices of the different materials. The design that is most eco- 
nomical in locations where stone and sand are cheap, and steel 
expensive, will not be economical where steel is relatively the cheaper 
material. In the first case the use of deep beams with comparatively 
little steel will make for economy, while in the other case the mini- 
mum allowable depth will prove most economical. 

Items Affecting Economical Arrangement of Beams.—In the 
above discussion, the cost of concrete floor only was considered. 
Additional factors, some of which will be discussed below, often affect 
the economy. 

Cost of Columns and Footings——The spacing of beams in a panel 
affects the dead load of the floor system. ‘This, in turn, influences 
the sizes and cost of columns and footings. The cost of columns and 
footings will be smaller with close spacing of beams, since then the 
thickness of slab is smaller and the weight of the floor construction 
is reduced. In high buildings, the extra dead load may affect the 
economy of the design, especially when the size of columns is limited 
and, therefore, to carry the extra load, a large percentage of steel 
would have to be used. 

Sprinklers.—When sprinklers are used, their cost may also affect 
the economy of the arrangement of the beams. To make the sprin- 
kler system effective, it is necessary to place, in each line of sprin- 
klers, at least one sprinkler head in each bay, irrespective of the size 
of the bay, because, owing to interference of the beam with the 
play of water from a sprinkler head, the head in one bay has no 
effect on the adjoining bay. The number of sprinkler heads, there- 
fore, increases with the increase in the number of bays in a panel. 

Thus, the 20 by 22-ft. panel used in the example on p. 578 requires 
two lines of sprinkler heads. With a two-bay arrangement, two- 
sprinkler heads in each line per panel are sufficient, making in all 
four heads per panel. With a three-bay arrangement, three heads 
per line must be used, making a total of six heads per panel. The 
difference in the number of heads amounts to two heads per panel. 
Figuring the cost of sprinkler heads at $8, the saving per panel 
amounts to $16, or 3.6 cents per sq. ft. of the floor for the smaller 
number of beams. | 

Plastering.—When the ceiling is plastered, each additional beam 
in a panel adds to the cost of plastering, there is an increase not only 
in the area of the surface to be plastered, but also in the number of 
corners. Extra charge is made for each corner to be plastered. 


578 BUILDING CONSTRUCTION 


Thus, in 1925, in a 20-ft. by 22-ft. panel, the cost of two-coat plaster 
work, around New York City, would be $91.00 with two bays and 
with $110.00 three bays, or a difference of 4 cents per sq. ft. in favor 
of the smaller number of beams. 

General Suggestions as to Spacing of Beams.—lIf the cost of 
two types is nearly equal, preference should be given to the arrange- 
ment with fewer beams. The beams are then wider, and the slab 
thicker, so that the construction is more solid and is better able to 
withstand fire. Narrow beams and thin slab suffer much in case of a 
fire. 

In oblong panels, the beams should run in the long direction and 
the girders in the short direction, unless a different arrangement 1s 
required for other reasons. An exception to this is a case with very 
long spans, as shown in Fig. 578, d. ‘The formwork for the long 
beams would be costly, and therefore it is cheapest to place the 
girders on column centers the long way, and run intermediate cross 
beams. In oblong panels, one bay per panel will prove economical 
for short spans up to 15 ft. With such an arrangement, the beams 
running in the long direction frame into the columns and no cross 
girders are required. The introduction of an intermediate beam 
would require also cross girders, which are not needed with one-bay 
arrangement. 

Beams in Wall Panels.—The distribution of light is affected by 
the direction of the floor beams in the wall panels. Beams parallel 
to the window form a larger obstruction to the light than beams per- 
pendicular to the wall. For this reason, wherever the light is of any 
importance, the beams should be placed at right angles to the wall. 
Of course, this is not feasible with a design using intermediate beams, 
where the wall beam must be kept shallow to accommodate the win- 
dows. The intermediate beams would have to frame into the wall 
beams. Owing to limited depth of wall beams, it may not be possible 
to get sufficient strength to carry the intermediate beams. 


EXAMPLE OF BEAM AND GIRDER DESIGN 


The design of beam and girder floors is illustrated by the following 
example: . | 
Example of Beam and Girder Design 

Example 1.—Design a typical panel, consisting of slab, beam, and girder, 
for a reinforced concrete building to carry a live load of 200 lb. per sq. ft. The 
spacing of columns is 20 ft. by 22 ft. on centers. Granolithic finish, one inch 
thick, applied separately, will be used. 


a a; a ee 








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(To far page 678) 





ae 






Lage of colur 


Slab sl smc same THK Bor! rd bars. 16:0" — 
as interior pane/ Bent 10"0c 


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ee Oe i ee aoe Be 
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legative bending momen 
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oy ‘Moment curves ford 
seeing supported and conti 
beam (Parallel line 


Points where bars 









Dash lines denote limiting 
points for bending bars aia Limiting pos ot hie a 
ae, Sq. 70 ae — fut Teo Stirrup Support _ | 












Sli In 741A 









ain 2-159, 2620'Str | /st bar ly 
‘ ‘ UO) wp ON imiting points 
ETRE SIAN icity ge 1S BERS td pe 
8 pe 5 sy pg compression bars gt" i oke4 hen g> 
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Beam B3 Beam By. compression bars 


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er rE NEES inte Pee ye ee 


2=1"59. 25'-0"Str 


1 3p 2'9 2640" 
1~1'Sg. 29'-10"SK#1 = SLIPR BEAN SHS 
1-1°Sg. 29'-10"Sh#2 “J-1"$ B5'-4" Skt Mes ap Boris 
8-35 gf 5-5 Ok#8 32 g S'S Sk#8 In all beams same as 
Beam Ba Beam Bo_ shown in Bt 


Fig. 189.—Details of Design 


35500 

















Max.v=——-———_- 
4 yA 212000. sino ys eee en 
S a a am i ' 
. 14x 091x225 Min.y =—8£/39 _.- 40, 
BS cs 1/4" Edge of colurings on” “Tax o9ex 265 7100/4 
is g Gent bars 1-1"Sq. each 63 60 
Stirrups F "ha I. 
3S 
| &3 .COre 
-Stirrups = £ of girder 
: 
ne { A 
\¢ of foam y 7 AS i edge of beam 
se: ee ee Papen STE 
Bent bars><-*> ~Stirrups Stirru, pss ——Bent bars= rem, Stirrups 
| ; y ‘lf 
0. LS pephbe ued i= Net span fessiled eg: 9 ais 





J 5 iG r / laa 
hear diagram for beams Shear diagram for girder 





aces for bending down bars 
I" for Ist. bar 


or 2nd bar 
or grd bar P 
Bp for #04 bar | =. Ss eel C= apebrte a= cat 





€--Moment curve|at support bd Out Rivho sf 
a 4 


Rririee wien elle bel 7 

Pe 1-11" | 20" 15:4" | 204 7-.2"|7922"| 2-4" 
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pri 0" —| 10-6" 24" 8.4” eceeae 2:10" 
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edocs eee 


nent Curve for Girder 6; _ wa sketches for bars 















aight 
—s 
a 

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RN 


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Cont. Girder 


Positive B. moment 


g 
S 
Ss 

Posite B moment jor 

simply supported girder 

















Bars from Ad- 
( yoining Span 


i Bent Bars 


limiting points for 4 
Kite? eee sia : ft. ire = ik 



















aa ag 
ay ie Peo 
the aig" Limiting points. peony am 
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=o * compression T= 1° D 350" SKES 
' oe 0" "SRO 


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= 
Dora ose G, 





Typical Panel. (See p. 579.) 
(To face page 579.) 


















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bear Sy 5 


DESIGN OF BEAM AND GIRDER FLOOR 579 


Specified unit stresses in lb. per sq. ft. are as follows: f, = 800, fs = 16 000, 
n = 15, v = 40 without stirrups, » = 120 with stirrups, « = 80 for plain bars 
and 100 for deformed bars. At support of continuous beams, f, = 900. 

Solution.—The girder is placed in the short direction .and the beam in the 
long direction, as shown in Fig. 189, opposite p. 579. 

Constants.—The following constants, taken from table on p. 205, correspond 
to the assumed working stresses: 

Neutral axis and moment arm, k = 0.429,7 = 0.857. 

Constants for depth of rectangular beam and slab respectively, C = 0.083, 
C, = 0.024. Ratio of steel for balanced design, p = 0.0107. 

Bending Moments and Span Length.—Bending moments recommended on 
p. 278 will be used. Span lengths are recommended on p. 277. 


Slab. Loads: Live load, 200 lb. 
Slab load, 56 lb. (assumed) 
Granolithic finish, 12 lb. 


Total, 268 lb. per sq. ft. 
20.0 
Gross span oe ae 10 ft. O in. 


Assuming width of beam equal 1 ft., net span / = 10.0 — 1.0 = 9 ft. O in. 
Using formula M = ;5wl? ft.-lb. or wl? in.-lb. (p. 279). 


M = 268 X 92 = 21 700 in.-lb. 
Depth of slab from formula d = Ci+/M (p: 208) where C; = 0.024 
d = 0.024+/21 700 = 3.5 in. 


Using #-in. concrete below steel, the distance to center of bars will be 1 in. 
and the total depth h = 3.5 + 1 = 4.5 in. 

Area of steel from formula A, = pbd. 

A, = 0.0107 X 3.5 = 0.088 sq. in. per inch. 

The spacing of 3-in. rd. bars is obtained by dividing area of one bar by 0.038. 


0.196 
It is —— =5.15in. Use ?-in. rd. bars 5 in. on centers. 
0.038 


The slab steel will consist of alternate straight and bent bars. The straight 
bar will extend over two bays and will be 20 ft. 6 in. long. The bent bars will 
serve as positive reinforcement in one bay, while the ends will be bent up and 
extended into the adjoining panel beyond points of inflection. (See plan.) 

Beam B,.—Beams B, extend between girders, therefore Formulas (66), p. 279, 
will be used. 

Span, center to center, 22 ft. Assuming width of girder equal 1 ft., net span, 
22 —1 = 21ft. Spacing of beams, 10 feet. 


Loads: Slab load 268 «K 10 = 2680 lb. 
Beam, below slab, 300 Ib. (assumed) 





2980 lb. per lin. ft. 


580 BUILDING CONSTRUCTION 


For intermediate beams spanning between girders, use: 
M = +,ul? ft.-lb. or wl? in.-lb. for negative and positive bending moment. 
(See p. 279.) 
UM = wil? = 2980 X 21? = 1315 000 in--lb. 
External shear, 


V = 22 x 2980 = 31 300 lb. 


Breadth of Flange.—Breadtk. of slab equals either 16 times the thickness of 
slab plus the breadth of stem of beam, or 4 of span of beam, whichever is smaller. 
Assuming b’ = 12 in., b = 16 X 4.5 + 12 = 84 in. In this case, } of span of 
beam is smaller than the above value; therefore the breadth of flange will be 
made equal to + X 21 X 12 = 63 inches. 

Cross Section as Determined by the Shear. 


V = 31 300 lb. 
b’ (2 _ 5) = SS = 260 sq. in. (see p. 218) 
Depth Required 
tee t by Shear 
b’ 2 d 
12 in. 21.6 23.85 
14 in. 18.6 20.85 
16 in. 16.2 18.45 


Minimum Depth at Support.—In continuous beams, the beam at support isa 
rectangular one with steel in top and bottom. The depth should not be smaller 
than required at support. In this case, it is desired to bend up one-half of the 
reinforcement and use the other half as compression reinforcement. The avail- 
able amount of compression reinforcement, therefore, is equal to one-half of the 
tensile reinforcement, or p’ = 3p1.. Assume d = 0.1. Since the allowable stress 

iS 16 000 


b tf. = 900, the ratio == — 
at support f, e ratio 15f, 15 X 900 


= 1.18. In diagram for a = 0.1 


on p. 908, using line a = 1.2, it is found that for pi: = 0.0182, p’ = 0.0091.* 
Cc 


This gives the required ratio between the compression and tension steel. The 
minimum depth, then, from Formula (26), p. 222, assuming b = 14 in. 


a= 1.054] 1 315 000 oye 
Aine 14 0.0182 X- 16.000 ames 





*In the diagram is drawn a straight line giving the required relation between the tensile steel 
and compression steel by connecting any two points for which p’ = 3p;. The intersection of 


this line with the line corresponding to eo = 1.2 gives the required ratio of tensile and compres- 
OSC 


sion reinforcement. For any other ratio a similar line may be drawn. Instead of the actual 
drawing of a line, the intersection may be obtained by using a straight edge. 


DESIGN OF BEAM AND GIRDER FLOOR 581 





rM st 
Economical Depth of T-beam.—From Formula (18), p. 216, d = “f° 5 + 5: 
if the ratio of unit cost of steel to cost of concrete is r = 70. 


for b’ = 12 d = 21.7 + 2.25 = 23.95 
b’ = 14 d = 20.0 + 2.25 = 22.25 
b’ = 16 d = 18.7 + 2.25 = 20.95 


Minimum Depth in the Center.—This depth is governed by the compression 
stresses in the center, where the beam is a T-beam. 
In this case the minimum depth is such that the neutral axis is within the 


a 
flange. Therefore ordinary beam formula, d = c,| >? is used. 


For fe = 800, fs = 16000, n = 15, C = 0.083. 
Since M = 1315 000 in.-lb. and b = 63 in., the required depth is 


1315 000 
nce vieseeg aa 1) 8 in, 


Depth Finally Selected—By inspecting the various depths computed above, 
the following values will be selected: 


b’ = 14in., d = 22.5in,, h = 22.5 + 2.5 = 25 in. 


This is the economical depth. It is larger than required at the support and some- 
what larger than required by shear. 
Area of Steel. 


t 
For d = 22.5 in., ¢ = 43 in., ade 0.2, the ratio of moment area is j = 0.91. 
The area of steel, then, is, from Formula (19), p. 216, 


1315 000 


As; = 
0.91 XK 22.5 X 16 000 





= 4.0 sq. in. 


. Use four 1-in. sq. bars with an area A; = 4 X 1 = 4.0 q. in. 

One-half of the bars will be carried straight and the other half will be bent up. 
The bars will be arranged in one layer. To distribute the steel properly, the bars 
will be bent up at two points at each end. The points of bending are taken from 
Fig. 100, p. 294. . 

Bond Stresses at Supports.—Since V = 31 300 lb. and the perimeter of bars is 
Zo = 4X 4 = 16 sq. in. j = 0.91, d = 22.5, the bond stresses are: 


31 300 
UU = 
0.91 X 22.5 X 16 





= 95 lb. 


This is satisfactory for deformed bars. 

Diagonal Tension Reinforcement.—Unit shear is plotted in Fig. 189, opp. p. 579. 
As it is desirable to utilize the bent bars for diagonal tension reinforcement, their 
location is marked on the shear diagram. As explained on p. 250, the bent bars 
may be considered as effective for a distance equal to d, provided their area is 


582 BUILDING CONSTRUCTION 


sufficient to take the stresses. The area tributary to each group of bars is marked 
on the diagram. 
Section ab. In this section stirrups must be provided. Shear to be resisted 
69 + 57 : ’ ; 
equals aa x 14 X 12 = 10600 lb. The area of } in. rd. stirrups with 
two legs equals 0.196 X 2 = 0.392 sq. in. and the available strength equals 


10 600 
392 X 16000 = 6300 lb. The required number of stirrups equals 6300 





= 1.7. 


Use 2-4 in. rd. stirrups in section ab. 

Section be. This section extends 3d on each side of the bent bar, therefore, 
the bent bar is effective in this section. 

Shear to be resisted 


57 -+ 39 
a x 14 X 22 = 14800 Ib. 


Stress to be resisted by bent bar 0.7 X 14 800 = 10 400 Ib. 

Area of bent bar 1 — 1 in. sq. bar = 1.0 sq. in. 

Available strength of bent bar 1.0 X 16000 = 16000 lb. The available 
strength is larger than the stress to be resisted. 

Section cd. The bent bar effective in this section is the same as in section be. 
Since the shear is smaller than in the other section, the bent bar is sufficient to 
resist the stresses. 

Section de. ‘The shear in this section is small. Two 3 in. rd. stirrups will be 
used arbitrarily placed as shown in the diagram. 

Beams B).—The unit loading and gross span for beams B, are the same as 
for beam B,, but their net span is smaller and the condition at the support dif- 
ferent. Beams B, run into flexible girders and therefore must be considered as 
continuous beams designed by Formulas (63) to (68), p. 278. Beams Bz, 
on the other hand, since they run into stiff columns, may be designed as building 
frames by Formulas (69) to (74), p. 279. 

Net span of Beam B, is 1 = 22 — 2 = 20 ft. 

Unit load 


w = 2980 lb. per lin. ft. 


Negative bending moment (Formula (69), p. 279). 
M, = + wi? = — 2980 < 20? = 1 192 000 in.-lb. 


Positive bending moment (Formula (70), p. 279). 


M = 0.75 wil? = 0.75 & 2980 X 202 = 894 000 in.-lb. 
Shear 
V = 2980 X 2,2 = 29 800 lb. 


The bending moment for this beam is smaller than for beam B;. Usually 
it is desirable, for the sake of appearance, to use same size for beams B, as for B,. 
If not, the economical dimensions may be found as explained in connection 
with By. 


DESIGN OF BEAM AND GIRDER FLOOR 583 


Using same dimensions as for B,; 
b= 14'm. a. ==,22,5 in. Py Rapes, Bis 


The areas of steel required by bending moments are: 
At support 
1 192 000 


a Seah 
10.91 X 22.5 X 16 000 ana 





In the center of span 





ak 894 000 Bg ae 
eer 0162285446 000 iL. ture ten 
Use in the center 2-2” round straight bars = 1.20 sq. in. 
2-1” round bent bars = 1.5/ sq. in. 


Toul 2.77 sq. in. 
This area is satisfactory. 

At the support the available reinforcement consists of two bent bars plus 
two bent bars extended from the adjoining span, making 4-1” round bars in all 
with an area of 4 X 0.785 = 3.14 sq. in. The available area of bent bars is 
smaller than the required area. The difference 3.64 — 3.14 = 0.50 sq. in. is 
supplied by a straight bar. 

As a result, reinforcement at the column consists of 


4-1” round bent bars = 3.14 sq. in. 


Pay 


4’ round straight bar = 0.60 sq. in. 


Total 3.74 sq. in. 


Check of Compresson Stresses at Support.—In this case the dimensions of the 
beam as well as the amount of tension and compression reinforcement are fixed. 
If the 2-2” round straight bars at the bottom of the beam are extended suf- 
ficiently into the columns to develop them by bond, they may be considered as 
compression reinforcement. 

The compresson stresses will be investigated 


b = 14 in. ye eae bien ld e225 =) sain: 
; { 3.74 
Tension steel A, = 3.74 sq. in. pi = 315 = 0.0119. 


1.20 
Compression steel A’; = 2 X 0.60 = 1.20 sq. in. p’ = ae 0.0038. 


The top bars are covered by concrete 1} in. thick. The distance from top to 
center of compression steel is 


= 0.089. 





ad ae + (7x4) =2in. t+ anda ity" 


584 BUILDING CONSTRUCTION 


15fe 


to p: = 0.0119 and p’ = 0.0038, which is ia = 1.52. When steel is stressed 





Using diagram on page 908 for a = 0.08, find the value of corresponding 





16 000 
in tension to f, = 16000 lb., the expression changes to 


= 1.52, and 
15fe 


the corresponding stress in concrete is 


16 000 


gee cet cteten hm aye er 
FRSC Edn Pe a 


Stirrups.—Since the external shear of Beam Bz is practically equal to Bi, the 
same spacing of stirrups will be used for all beams. 


Girder G:.—Gross Span 20 ft. 0 in. Assuming that the minimum size of 
column is 18 in., the net span is 20.0 — 1.5 = 18.5 ft. 


Loading.—Concentrated load 2980 X 22 = 65 500 Ib. in center of span. 
Weight of girder (below slab) 300 lb. per lin. ft. 


Bending Moments. 
Concentrated loads, considering the span as simply supported. 


65 500 _ 18.5 
M = eae x Ys < 12 = 3 640 000 in.-lb. 


For continuous span, multiply by a ratio of 3- 


M = 3640 000 X 58; = 2 430 000 in.-lb. 


For uniform load, 


M = 300 X 18.35? 


103 000 in.-Ib. 
Total 2 533 000 in.-lb. 


This bending moment will be used in the center and at the support. 


External Shear. 


65 500 
LABS = 32 750 lb., concentrated load 
18.5 : ae’ 
V2 = 300 X re 2770 lb., uniformly distributed load 
Total V = 35 520 lb. 
Area required by shear, | 


t 35 520 
v( ~ s) = ——— = 296 sq. in. 


DESIGN OF BEAM AND GIRDER FLOOR 5085 
Minimum dimensions required by shear, 


b’ = 14 in. d — 21.6 in. 


b’ = 16 in. 1S 


5 19.0 in. 


Minimum Depth at Support.—Since the ratios of steel in tension and com- 
pression for minimum depth for the girder are the same as for the beam Bi, 
namely, pi = 0.0182, p’ = 0.0091, the minimum depth at support 


= 1.05 4| 2 533 000 ee 
ieee 14.500. 018290 16 000) cs 





This depth will be accepted. It is larger than required by shear. 
Area of Steel in the Center of Span. 


t 
Ford = 26.2in.,¢ = 43 in; = 0.17, from table on p. 221, the ratioj = 0.92. 


The area of steel, then, is 
2 533 000 


Dg 
0.92 XK 262 X 16 000 


= 6.57. 


Use 6-1" rd. 4.71 sq. in. 3 straight 3 bent 
2-1” sq. 2.00sq. in. 1 straight 1 bent 


Total 6.71 sq. in. 


The bars will be placed in two rows. They will be bent in two places. To 
find the points at which to bend the reinforcement, the bending moment diagram 
must be drawn, as explained in Vol. III. The variation of bending moments for 
concentrated loads is different from the variation for uniform loading, therefore, 
the diagrams on pp. 292 to 297 do not apply. Since the bending moment due to 
the dead load is small, the same variation of the total bending moment may be 
taken as for the concentrated load alone. The bending moment diagram and the 
bending diagram for reinforcement are shown in Fig. 189, opposite p. 579. 


Bond Stresses, V = 35 500 lb., jd = 0.92, d = 25.8. 


Zo = 6X 3.1442 X 4 = 26.8 in. 
Hence, 
35 520 


Ug eee oe = 65 Ib. 
0.92 X 26.2 XK 26.8 


Bond stresses are satisfactory. 


Diagonal Tension Reinforcement.—The shear diagram is drawn first. (See 
Tig. 189.) The location of the bent bars is then shown and the area tributary 
to each bent bar is marked off. The shear diagram is then divided into several 
sections. Each section is investigated as given below. 


586 BUILDING CONSTRUCTION 


The shear diagram is divided into sections ab, be, cd, de, and ef. The stresses 
in each section will be computed and the required reinforcement provided. 
Section ab. In this section stirrups must be provided. Shear to be resisted 
68 + 67 





equals x 14 X 6 = 5700 Ib. Since the strength of 3 in. rd. U-Stirrup 


equals 0.392 X 16 000 = 6300 lb., one stirrup is sufficient. 
d 
Section be. This section extends 5 on both sides of the bent bar. The bent 


bar, therefore, is fully effective in this section. The diagonal tension to be 
: : 67 + 65 ‘ 

resisted is (v — v’)bs = oe < 14 xX 27 = 25000 lb. This produces stresses 

in bent bars in this section equal to A;f; = 0.7(v — v’)bs=0.7 XK 25 000 = 

17500 lb. (See Formula (39), p. 248, also p. 155.) Since in this section two 

1-in. rd. bars are bent with an area 0.785 X 2 = 1.57 sq. in. and a capacity 

1.57 X 16.000 = 25 100 lb., the diagonal may be considered as taken care of. 


65 + 63 
65 +68 V4 x 27 = 24200 1b. 





Section cd. The stress to be resisted equals 


The stress produced in a bent bar equals 0.7 X 24 200 = 16900 lb. In this 
section 1-1 in. sq. bar is bent up with an area of 1 sq. in. and capacity of 16 000 lb. 
This is smaller than 16 900 lb., therefore, the bar is not sufficient to take care 
of the diagonal tension stresses. The difference 16 900 — 16 000 = 900 lb. 
Divided by 0.7 this gives the stresses to be provided for by stirrup 900 + 0.7 
— 1290 lb. One } in. rd. U-stirrup in this section gives sufficient area. 

Section de. The stress to be resisted equals are xX 14 XK 27 = 23 300 
Ib. In this section 1-1 in. rd. bar is bent up. Its capacity equals 0.785 X 
16000 = 12600 lb. This divided by 0.7 and subtracted from the total 
stress to be resisted gives the stress to be taken care of by stirrups. 23 300 — 
(12 600 + 0.7) = 5300 lb. This stress may be added to the stresses in section ef. 
noe x 14X14" "1T 700 Tb. 
Adding to it 5300 lb. from the previous.section makes the total stresses 17 000 lb. 
This requires three 3 in. rd. stirrups. 

With the above figures as a guide stirrups are spaced as shown in the figure. 
One stirrup is placed on each side of the beam to take care of the load transmitted 
by the beam. Three additional stirrups are placed between the beam and the 
first bent up bar. The number of stirrups used near the column is larger than 
actually required. 


Section ef. The stress to be resisted equals 


END PANEL 


Slabs.—Slab in the end panel is of the same design as for interior panel. 
Beam B;.—The net span and loading is the same as for corresponding interior 
beam, namely Beam Bj. 
b 2d fu. 


w = 2980 lb. per sq. ft. 


FLAT SLAB FLOOR CONSTRUCTION 087 


Since beams B; run into girders, they will be designed as continuous beams by 
means of Formulas (63) to (68). 


Negative Moment 


At Interior Column M, = — 1.2 wl? = — 1.2 X* 2980 & 212 = 1580 000 in.-lb. 


At Wall Column WM, = — 0.6 wi? = — 0.6 X 2980 X 212 = 790 000 in.-lb. 
Positive Moment is same as negative moment at interior column 

M = 1 580 000 in.-lb. 
Use same dimensions as for interior span 


Do 14. in. @ =. 22-5111. hp e250. 


; M 
The required area of steel is found from As; = af, 
JIGS s 
In the center and at interior column 
1 580 000 : 
8 = 4.82 sq. in. 





~ 0.91 X 22.5 X 16 000 


At the wall column the moment is equal to one-half of the moment in the center, 
therefore, the required area of steel 


Asi a 7A 


I 


2.42 sq. in. 


Use in the center 5-1” rd. bars = 3.92 sq. in. 
doi eeqs bar) )'="1),00 sq.in. 


Total 4.92 sq. in. 


Bend up 1-1” sq. bar and 2-1” rd. bars, which give an area of 1.0 + 
1.57 = 2.57 sq. in. The bent bars extending from the adjoining interior span 
equal 2-1” sq. bars with an area of 2.0 sq. in. The total area, therefore, is 
2.57 + 2.0 = 4.57 sq. in. Since the required area is 4.82 sq. in., the balance 
4.82 — 4.57 = 0.25 sq. in. must be supplied by means of straight bar plaed 
on the top. In this case 1-3” rd. bar will be used. 

The bars are arranged in accordance with diagram on page 293. 

The reinforcement at the wall end of the beam is clearly shown in the eleva- 
tion of the beam. 

Stirrups.—Same stirrups will be used as for corresponding interior beam. 


FLAT SLAB FLOOR CONSTRUCTION 


Flat slab floor construction is used to a great extent in building 
construction. It is particularly adaptable to warehouses, industrial 
buildings, garages, and automobile service stations, where it is cheaper 
than any other type of floor construction, 


088 BUILDING CONSTRUCTION 


In buildings with large spans and light live loads, flat slab con- 
struction may not prove economical, especially when column capitals 
are not permitted. 

For very long spans, over 35 ft., the dead load of flat slab may be 
too large. 

Advantages of Flat Slab Construction—Following are the 
advantages of flat slab construction over beam and girder con- 
struction. 

1. Lower cost (see comparison of cost on p. 788). 

2. Smaller depth of construction which results in saving in height 
of building. 

Under ordinary conditions the saving in height in a ten-story 
building over beam and girder construction is sufficient to add 
another story. This probably is of particular interest in cities, where 
the height of buildings is limited. 

3. Flat ceilings give better distribution of light. 

4. Due to absence of beams and corners in columns, flat slab 
construction resists fire better than any other type of concrete. 

5. Greater efficiency of sprinklers, because there is no interference 
with the play of the stream of water. 

Design of Flat Slab Construction—Complete information regard-_ 
ing the design of flat slab construction is given in Chapter VI. 


LIGHT-WEIGHT FLOOR CONSTRUCTION 


Under this heading come floor constructions in which the solid 
concrete slab is replaced by closely spaced concrete ribs connected 
at the top by a thin concrete slab, usually called topping. The space 
between the joists below the topping is either filled by tiles, or is 
hollow. Such floors may be divided into the following groups, 
according to the method used for the elimination of the tension con- 
crete between the joists. 


1. Hollow tile concrete floors. 
2. Metal tile concrete floors. 


In some cases, the metal tile is replaced by removable wood forms, 
as in the Bransom System. ‘The design of this floor is the same as for 
metal tile. The forms are described in Volume II of this treatise. 

Use of the Light-weight Floors.—This type of construction may 
be found economical where light floor loads are carried on long spans, 
and where smooth ceilings are required. It is used to a great extent 


LIGHT-WEIGHT FLOOR CONSTRUCTION 589 - 


in schoolhouses, hospitals, and office buildings, in combination with 
either reinforced concrete or structural steel frame. It is not eco- 
nomical for heavy loads. Since the topping is thin, it should not 
be used where there is a possibility of concentrated loads. 

The light-weight construction is lighter than the solid concrete 
slab construction. This reduces the dead load on the girders and 
columns. The actual economy of the construction will depend upon 
the relative cost of the concrete and the tiles. In determining the 
relative economy of a solid slab floor and a metal tile floor, in cases 
where flat ceiling is required, it is necessary, to add to the cost of the 
tile floor the cost of the metal lath which is required in metal tile floors 
for suspended ceiling, also the difference between the cost of plas- 
tering on metal lath and on concrete. 

Comparison of Various Types of Light-weight Systems.—In 
deciding on the type of light-weight floor, the cost of the system, as 
well as the adaptability, should be considered. The hollow clay 
tile floor is better adapted for heavy loads than the metal tile floor, 
because the joists are spaced closer and therefore have larger shearing 
strength for the same depth. It is also more fireproof, as only the 
bottom of the joist is exposed to the fire. In metal tile floors, if no 
suspended ceiling is used, the narrow joist and the thin slab are 
exposed to the fire on all sides. The resistance of such floors to fire is 
much smaller than that of solid concrete floors. 

When the ceiling is not plastered, the appearance of hollow tile 
floor is not pleasing. Metal tile floors built with removable forms, 
may be used without plastering. 


REINFORCED CONCRETE HoLLow TILE FLOOR CONSTRUCTION 


For long spans, the weight of the floor construction. may be 
reduced by the use of a combination of hollow tile and reinforced 
concrete, in which hollow tile replaces the concrete in the tensile 
portion. The tile is lighter than the replaced concrete; therefore, 
the resulting construction weighs less than a solid concrete slab. 
The combination of hollow tile and concrete may be considered as 
intermediate between solid concrete floor and metal tile floor. It is 
considered by the authors as the best light-weight type. 

General Description.—The most common combination of hollow 
tile and concrete is shown in Fig. 190, page 591. ‘The construction 
consists of hollow tile, reinforced concrete joists poured between the 


590 BUILDING CONSTRUCTION 


tiles, and concrete topping above the tiles, poured monolithic with 
the joists. The joists may run in one direction only, in which case 
the combination is called one-way hollow tile system (see p. 591), or 
they may run in two directions at right angles to each other, the 
construction in this case being called two-way hollow tile system 
(see p. 592). The joists are supported by reinforced concrete girders, 
by structural steel beams, or by walls. In the two-way system, sup- 
ports must be provided on four sides. 

The tile may be of burned clay, gypsum, or any other material 
adaptable for the purpose. 

Hollow Clay Tiles.—Burned clay tiles are most commonly used. 
Figure 191 illustrates a typical hollow floor tile made of fire clay and 
used in the one-way system. In plan it is 12 in. square and varies 
in depth from a minimum of 4 in. to a maximum of 15 in. The sur- 
faces of the tiles are usually scored, in order to bind the tile and the 
concrete more thoroughly and also to furnish a good plastering sur- 
face. When an all-tile ceiling is desired, special 1-in. tile slabs are 
placed between this under the joists. While this has the advantage 
of providing a uniform plastering surface, it increases the depth of 
construction by 1 in. 

The weight of the tile depends upon the material of which it is 
made and also upon the thickness of the walls of the tile. The 
weights given in the table below are for burned clay tiles as manu- 
factured by the National Fireproofing Co. 


Weight of Hollow Burned Clay Tile 


aa aan Le en ee 














Size of Tile Weight Size of Tile Weight 

in Inches per Tile in Inches per Tile 
D2 Wek 2 16 lb. | 8 X 12 X 12 32 Ib, 
3X12 412 16 lb. . (Xe 36 Ib. 
4 Xap s 18 lb. 10° * 125633 38 Ib. 
5 xX 12x 12 QT oe 12 ie 42 lb. 
a Poul A i be 24 Ib. 15 X12 %12 50 Ib. 
TR ALRG 12 29 Ib. 











Se 


In the one-way arrangement, the clay tiles are placed with ribs 
parallel to the direction of the joists. When the tiles are sound and 
are properly placed, comparatively little concrete finds its way inside. 


591 


LIGHT-WEIGHT FLOOR CONSTRUCTION 





ee Se 


N/ 


~ 


| 
| 
| 
| 
| 
| 
| 
| 
| 
| 
| 
| 
si 
| 
| 
| 
| 
| 
I 
| 


4 


Stag apart eo eae 





Section Through Joists 


ULE ee STs. 





=| 
S's 
= 
Ss }- 
8 
33 | 
Laika) 
aie) 


Section Through Girder 


(See p. 590.) 


Fic. 190,—Hollow Tile and Concrete Floors. 


592 BUILDING CONSTRUCTION - 


When the tiles are broken or when the joints between the tiles are 
not properly sealed, considerable concrete may flow into the tiles, 
thereby increasing not only the weight of construction but also its 
cost. 

In the two-way arrangement, some designers use regular tiles, 
claiming that no considerable amount of concrete flows into the tiles 
through the open ends. The authors do not consider such designs 
desirable, because it is impossible to estimate the increased weight of 
slab and amount of extra concrete. Special tiles, for the purpose 
of sealing the open ends, are on the market. In the arrangement 
shown in Fig. 192, p. 593, each group between four joists consists 
of six tiles, forming a 24-in. 
square. With 4-in. joists, their 
spacing in both directions is 28 
in. on centers. 

In another arrangement, the 
ends of the tiles are closed by 
means of channel tiles, the 
flanges of which are placed 
against the open ends of two 
adjoining tiles. The web of the 
channel is placed on the form, 

Fic. 191.—Typical Hollow Clay Floor and its width fixes the spacing 
Tile. (See p. 590.) of the tiles. The width of the 

joist is equal to the clear dis- 

tance between the flanges of the channel. Additional soffit tiles are 
used, and an all-tile ceiling is thus obtained. (See Fig. 201, p. 613.) 

Gypsum Tiles.—Hollow tiles may also be made of gypsum. The 
standard width of the gypsum tile is 20 in.; the spacing of 4-in. 
joists would therefore be 24 in. on centers. 

Gypsum tile is lighter than clay tile. An additional advantage 
is that it can be readily cut in the field. It has the disadvantage, 
as compared with clay tile, that it is much more brittle and therefore 
requires more careful handling. 

Joists.—The joists may be made of any width, by spacing the 
rows of tiles the desired distance apart. The minimum width of the 
concrete joists should be 4 in. This width has been adapted as 
standard by many users. A smaller width is sometimes used with 
burned clay tiles, but then the space is filled with cement mortar 
and not with concrete. 





















































LIGHT-WEIGHT FLOOR CONSTRUCTION 593 


Topping.—The concrete topping over the tiles should be at 
least 2 in. thick, and should be poured in one operation with the 
joists Since the concrete topping rests on the tiles, no bending 
stresses can be developed in the concrete and no bending reinforce- 
ment is necessary. It is advisable, however, to use some shrinkage 























Fig. 192.—Hollow Clay Tiles for Two-way Combination Floor. (See p. 592.) 


and temperature steel to prevent cracking of the topping. The 
amount to be used can be smaller than that required in metal tile 
slabs, as the contraction of the concrete is reduced somewhat by the 
adherence of the concrete to the tile. 

In clay tile construction for light loads, the topping is sometimes 
omitted and the construction then consists of rows of tiles separated 
by rectangular concrete joists of the same depth as the tiles. Nat- 


594 BUILDING CONSTRUCTION 


urally, the tiles must be strong enough to carry the load between the 
jOistseiee 

Design of Joists.—The floor loads are carried by the concrete 
joists, which, with the topping, form small T-beams. The design of 
the floor, therefore, resolves itself into the design of small T-beams. 
Each joist carries the dead load plus the live load of a strip of floor of a 
width equal to the spacing of the joists. The bending moments and 
shears are computed in the same manner as for regular concrete 
beams. The bending moment coefficients will depend upon the 
degree of restraint at the supports and upon the number of spans. 
The span to be used in design 1s discussed on p. 277. 

For simply supported floors, such as’ those in which the joists 
rest on brick walls, it is necessary to compute shearing stresses at the 
support, the amount of steel in the center, and the compression 
stresses in the center. The economical depth and the depth required 
by deflection are usually larger than the depth governed by the com- 
pression stresses. 

In continuous beams, in addition, the amount of steel and the 
compression stresses at the continuous support must be found. At 
the support, the topping, which in the center forms the flange of the 
T-beam, is in the tension zone, so that the joist becomes a rectangular 
beam. Usually, the compression stresses are high and compression 
steel is required. ~The depth required at support may be the govern- 
ing depth. , 

Shearing Stresses.—Shearing stresses involving diagonal tension 
should always be computed. If it is desired not to use any diagonal 
tension reinforcement, the unit stresses must not exceed the value 
allowed for plain concrete. Many building codes require that, in 
computing shear, only the width of the concrete joist shall be con- 
sidered as effective. At present, there is a tendency to make some 
allowance for the resistance of the clay tile to diagonal tension. The 
report of the Joint Committee of 1924 gives the following rule, 
which is endorsed by the authors: 3 

“Shear in Beam and Tile Construction.—The width of the effective 
section for shear, as governing diagonal tension, shall be assumed as 
web plus one-half the thickness of the vertical webs of the concrete 
or clay tile in contact with the beam.” | 

Reinforcement of Joists——The amount of reinforcement is 
obtained from the bending moments, as recommended on p. 275. 
In selecting the number of bars, the bond stresses must be taken into 


LIGHT-WEIGHT FLOOR CONSTRUCTION 595 


consideration. To keep the bond stresses within working limits, the 
use of bars of small diameter is recommended. Since the joists are 
narrow, the tendency is to use a small number of heavy bars. This 
produces excessive bond stresses. 

It is not advisable to use more than two bars in one row. If a 
larger number of bars are used, they should be placed in two rows, 
in which case proper allowance must be made in computations for 
the reduced moment arm. 

In continuous joists, it is advisable to use an even number of bars 
of nearly the same size. Bars forming about one-half of the total 
area of steel may be bent up and carried over the support. When 
the reinforcement of the joist consists of one bar only, it must be 
carried straight at the bottom for the full length of the joist. If it 
is not possible to divide the steel in half, the larger portion should be 
bent up. It is advisable to carry at least 40 per cent of the bottom 
steel straight to the support, and a larger amount is often required 
by compression stresses. The proper amount of steel should be 
provided at the support. The bent-up bars with the bars coming 
from the adjoining span are usually sufficient. If not, additional 
straight bars should be used. When no bars are bent up, the full 
area of steel required for negative bending moment must be provided 
by short bars. 

The bars should be bent up at such points that the center of the 
bend coincides with the point of inflection. They should be extended 
into the adjacent span a sufficient distance to develop the bar in 
tension for negative moment. ‘To serve this purpose, they must be 
extended at least 20 diameters beyond the points of inflection of that 
span. An attempt is often made to use the bent bars both as nega~- 
tive moment reinforcement and as diagonal tension reinforcement. 
The result is not satisfactory, as there is danger that neither of the 
stresses will be properly taken care of. At the end support, the 
proper amount of steel should be used to take care of any bending 
moments that may be developed. The magnitude of the negative 
bending moment to be taken care of will depend upon the amount 
of restraint. A minimum negative bending moment equal to ygwl* 
should be provided for. 

Points of Inflection.—The location of the points of inflection 
should be computed on the basis of the net span and not the gross 
span. The distance to the point of inflection should be measured 
from the edge of the support and may be accepted as 3 of the net 


696 BUILDING CONSTRUCTION 


1 
span for interior spans and 45 for wall panels. When the bending 


moment is computed on the basis of the gross span, there is the 
tendency to consider the point of inflection as distant from the center 
of the support a distance equal to + of the gross span. For wide 
supports, this assumption fixes the point of inflection much nearer 
the support than it actually 1s, with the result that the negative 
bending moment is not properly provided for. While the assump- 
tion of gross span in figuring the bending moment is safe, it gives 
unsafe results in figuring the distance of the points of inflection. 

Spacing of Reinforcement.—The bars should be placed not 
nearer to the sides of the tile than ? in. This 1s required for bond 
and not for fireproofing, as the tile serves as a sufficient fire protec- 
tion. The bars should be kept a proper distance above the form. 
When two layers of bars are used, they should be separated by a 
separator bar one inch in diameter. In narrow joists, it is difficult 
to keep the bars in place without mechanical spacers and chairs. 
In many instances, it has been found after the concrete was removed 
that the bars were misplaced, in pouring concrete, to such an extent 
that there was no concrete between the bars. 

Diagonal Tension Reinforcement.—If the unit stresses exceed 
the allowable values for plain concrete, stirrups should be used. 
The size and spacing should be determined as explained on p. 247. 
The bent bars may be considered as resisting diagonal tension in the 
portion of the beam where they occur. 

Selection of Depth of Joist——In selecting the depth of joists, not 
only the stresses, but also the deflection must be considered. For a 
joist that is shallow in proportion to the span, the deflection may be 
excessive even if the stresses are within working limits. If the ceil- 
ings are plastered, the deflection should not exceed 345 of the span; 
if it is greater the plaster will crack. 

Economy of design should also be considered in selecting the 
proper depth of joists. The economical depth will depend upon the 
relative cost of concrete tile and steel. By increasing the depth of 
the joist, the cost of concrete and tile is increased; but the cost of 
steel is reduced. A depth may be considered economical when any 
increase in it would give an increase in cost of concrete and tile greater 
than the accompanying decrease in cost of steel. 

In many instances, the depth required for compression at the 
support is larger than the economical depth. At the support, the 


LIGHT-WEIGHT FLOOR CONSTRUCTION 597 


joist is a rectangular beam with double reinforcement. The amount 
of compression steel may be assumed as one-half of the tension 
reinforcement, and the depth computed by means of the constants 
from Diagram, p. 908. The allowable compression stress at the 
support may be increased as explained on p. 282. 

Girders Supporting Joists.——The girders supporting the joists 
may be either of reinforced concrete or of steel. If reinforced concrete 
girders are used, the required flanges for the T-beams are provided 
by omitting the tiles adjacent to the edge of the girder. (See Fig. 
190, p. 591.) Asa result, a solid slab is obtained on both sides of the 
girder, of the same depth as the depth of the joist. The flange so 
obtained may be deeper than required to resist the compression 
stresses. If this is the case, instead of omitting the tiles altogether, 
thinner tiles may be used. For instance, in a floor consisting of 
10-in. tiles and 2-in. topping, by omitting tiles at the girders, tke 
thickness of the flange of the girder would be 12 in. In most cases, 
this is more than is required by the stresses. If 6-in. flange is required 
for the girder, instead of omitting the tiles at the girder, 6-in. tiles 
are substituted for the 10-in. tiles. As a result, the flange of the 
girder will be 6 in. thick. While this method saves some concrete, 
it may lead to confusion on the job. 

Support on Brickwork.—The joists should have a bearing on 
the wall of at least 6 in. The required area of bearing should be 
computed from the reaction of the joist and the allowable stresses on 
brickwork. 

Weight of Clay Tile Floors—In computing the dead load of 
the slab, the following weights may be assumed for burned clay tile 
with 4-in. concrete joist spaced 16 in. on centers and 2-in. topping. 
The weight of flooring and plaster should be added. If the thickness 
of topping is more than 2 in., the extra weight of concrete should be 
added. 


Thickness of tile, in... . 3 4 5 6 7 8 Diet LO ree Speen 
Amount of concrete per 
poettsin cu. tt....... 0.23 0.25 0.27 0.29 0.32 0.34 0.36 0.38 0.4 20.48 


Average weight of tile 
and concrete, per sq. 
OS See Ae ee a ae 45% 5D. 255, 1000.65) 702 ome OmenOU. LOG 


Erection.—It is evident that for one-way arrangement of tiles the 
form boards need be used only under the joists. The width of the 


598 BUILDING CONSTRUCTION 


boards must exceed the width of the joist by an amount sufficient to 
give support for the tiles. 

The tiles must be erected in a straight line and kept in place during 
concreting. Misplacement of the tiles would reduce the width of the 
joist and might have serious consequences. 

Tiles should be sprinkled with water before concrete is poured, as 
dry tile absorbs moisture from the concrete. If this is omitted, 
especially in dry weather, the topping will lose more water by absorp- 
tion than the joists, causing cracking and uneven setting of concrete. 

Since the joists are narrow, with a large amount of reinforcement, 
the large stones used as coarse aggregate would obstruct the flow of 
the concrete around the reinforcement and would form pockets. 
The maximum stone should therefore pass a {-in. sieve. The con- 
crete must be carefully spaded and worked around the reinforcement. 

The joists must be poured in one operation with the topping. 
Construction joints in the joist should be made in the center of the 
joist. Construction joints parallel to the joist should be made 
midway between the joists. It must be borne in mind that the 
topping acts as a flange for the joist, and nothing must be done which 
would interfere with this function. 

Example of Hollow Tile Floor.—Figure 193 gives the floor plan of 
a building in which hollow tile construction was used. The arrange- 
ment of panels and the sizes are given. ‘The computations for a 
typical panel are given below: 

Typical Design of Combination Hollow Tile and Reinforced Con- 
crete Floor.—The computations given below refer to the floor in the 
Ordnance Building shown in Fig. 193, p. 599.4 The floor is intended 
for office use. 


2 
Slab, Interior Span, 15 ft. 0 in—M = ae for positive and 
negative bending moments. 
Live:load 24.205. ae a eee 75 lb. per sq. ft. 
9-in. cinder concrete fill and sleepers.. 15 
T_in, wood flooring. ....--+ +++ +a 5 
1-in. ceiling plaster. 4) --...+--.943 S 
Assume weight. of slaboe. 2!) eee 60 
‘Total loads... sepa eae w, = 160 lb. per sq. ft. 


4 Arthur Wood, Architect, National Fireproofing Co., Engineers. 


LIGHT-WEIGHT FLOOR CONSTRUCTION 599 





O"/ong perth 
orst on 4h 


a, 7* 
ef 


Seat a ere Sek aati ee ee ———- — = — Ss ff rrr 


1-0" 


/ong per joist on 
top over Girder 


“ 


h. 
SS 

% 
> 

~y 
se 


JZ 


He BLOM ate EES 610" ake 


> "x6"Tile mulhons filled with conerete 
.f and eee cay, = 59. Bar Bar 








rad 6 











Reinforced concrete lintel 2-4 "Sa 
Plan of Triple Windows ras ie aie tee | Oe 


"Concrete top poured with the joists , 2” Cinder concrete 
Jill and, sleepers 
— = 













































ab id ee xl2*hollow tile 7 Fs ‘Woo ay ee | 
po SE EL wake aX = wr" 
1.000 al Bar we 8 
‘4 oy 
: 4-76" oe 3" -.t ie! pastas sIn Mullion 62+ 
nT oaks Section "A-A” -¢ Column Wall 8.—~ rk >| 
LS aoe on cae a ‘I sae Pe Pilaster 12"\~ 


yf section: C- ms 


as stirrups inverte 
Bh hc yal MCA 2 "Bars bent up 


These ae rups inverted. 
C73 j Bars pie alee || up 













ie Bal : 
End of bar froin 
adjacent span 
-- 3’-9"---> 


4-5 Fae 7 Bars pilaster reinforcement fi = 

cells of tile with concrete down to 

Joundation jor solid 
pilaster 










End of bar from 
adjacent span 















Letail Girder "G” 








Alternate Courses 


ft ers 


C L. of Girder-- 






















Typical Col Section 


Note:-/n alternate girders the# 
bars will be inside and thez’bars 
outside in order to permit bending a 
up of the f"bars in all girders. Cross-Section Girder "G 


Section’B-8” 
I'tg. 193.—Floor Construction in Ordnance Building, Washington, D. C. 


(See p. 598.) 


600 BUILDING CONSTRUCTION 


With joists spaced 16 in. on centers, the unit load per joist, 
= 160°* 42 = 213 Ibsperlinait. 
Assume odulin unit stresses: 


f. = 650 Ib. per sq. in. 
fs = 16000 lb. per sq. in. 
v= 40 Ib. per sq. in. 
u= 100 lb. persq. in. (for deformed bars). 


If'n = 15,9 = ? (approx.). 
The joists may be considered as T-beams. Their fixed dimensions 
are: 
Width of flange. ...b = 16 in. 
Width of stem..... b’ = 5in. (4in. concrete + 1 in. for tile shells) 
Thickness of flange. . = 23 in. (2 in. concrete + 3 in. for tile shells) 
The maximum shear at edge of girder. Net span equals 
15 — 1.5 = 18.5. 


y =e 140i 


The maximum bending moment, 


213 x (15)2 x 12 
12 


Use 6-in. tile with 2-in. concrete top (weight = 60 lb. per sq. ft.) 
and, allowing 1} in. below center of steel, d = 6 in. 
The required area of steel is, 


Mire 48 000 
f.jd 16000 X% X 63 
~ Use one 3-in. sq. bar in each joist (A, = 0.56 sq. in.). 

The area of steel was found for an assumed value of 7. The 
results will be checked by computing accurate values. Formulas 
on p. 134 are used. 

Check of Unit Stresses.—(Use formulas on p. 184.) 


QnAd+b2 (2X15 X 0.56 X 6.75) + 16 X (28)? 


M= = 48 000 in.-lb. 








Ane = 0.56 sq. in. 





kd = So > ax 15 X 050) Ce 
2.22 in. < 28 in.; therefore the neutral axis is in the flange. 
hale 2222 
= > = ae = 0.829, 
and 


4=1—- 3k = 0.890. 


LIGHT-WEIGHT FLOOR CONSTRUCTION 601 


Unit Stresses: 








M 48 000 

Je = Figo ~ FX 0320 X 0.800 X 16 X (G75)? — 70 IP. Per sa. in 
M 48 000 : 

ee tra 06x08 X6.75 Per aa 
V 1 440 





i = 48 lb. per sq. in. 


bid 5 X 0.890 X 6.75 
Stirrups are necessary at ends of beam. 


Mes 1 600 
weve ao < 0/890: X< 6.75 


Bond stresses are satisfactory for deformed bars. 

Joist at Support——Assuming the negative bending moment over 
interior supports to be equal to the positive bending moment at the 
center of the span, the same amount of steel will be provided as in 
the center of the span. One {-in. square bar will be used at the top 
of each joist over the girder, with effective depth of 6% in. from the 
bottom of the slab. These continuity bars will be made 7 ft. long 
to reach the quarter-points of the span each side of the girder. The 
positive reinforcement will be carried straight through the girder 
into the adjacent span, for a distance sufficient to take care of com- 
pression stress in the steel equal to the maximum stress as deter- 
mined when the joist at the support is figured as a double reinforced 


beam. Ce 
Unit stresses due to negative moment at support. (See p. 139.) 


= 89 lb. per sq. in. 





1+ 
6.75 
k = V2n(p + pa) + n2(p + p')? — n(p + p') = 0.418 
_ WL — 3k) + 2p’n{k — a) — 2) _ 9 ga 
a k2 + 2n’n(k — a) 
ia ape 48 000 
ad? 0.56 < 0.842'x'6.75 


= 0.185 





C= 





= 15 100 1b) per.sq.in. 


iF =f = 723 lb. per sq. in. (At support, a stress 


15 per cent higher than 650 is permitted = 750 lb., 
per sq. in.) 


fe =fs vo ib = 6 050 lb. per sq. in, 


602 BUILDING CONSTRUCTION 


Length of positive steel required beyond edge of support, as 
determined by bond for j-in. bar, 


fa 6050 x3 


= 11.4 in., say 12 in. 








4u «4. X 100 
Exterior Span.—Use the same slab construction for the two 
exterior spans of 14ft.0in.; M = i 
We LU = 1385 lb. 
2 ; 
yu = 218 x G4)" X 12 _'50 000 in.-Ib. 
10 
50 000 
i. = +0 309 X 0.590 X16 x 6.756) ay 
50 000 
fe = DFO NC0BIO KOI ee 
1 385 


v= 5 xX 0.800 X 6.75 = 45.2 Ib. per sq. 1. 


Stirrups at ends are required. 


i“ 1491 
~ 3 xX 0.890 X 6.75 


Bond stress satisfactory for deformed bars. 


U = 83 lb. per sq. in. 


STEEL TILE FLOORS 


Another type of light floor construction 1s the steel tile construc- 
tion. In this type the concrete in tension is eliminated by metal 
tiles of proper design. The resulting construction is shown in Fig. 
194, p. 603. These steel tiles are often called “ metal pans,” or 
““ tin pans. 

General Description.—Constructions of this type may be divided 
into one-way steel tile floors, in which the tiles run in one direction 
only, and two-way steel tile floors, in which the joists run in two 
directions at right angles to each other. In the one-way arrangement, 
each row of tiles consists of a number of tiles open on both ends, 
overlapping each other and provided at the ends with end caps. In 
the two-way arrangement, special tiles closed on four sides and on the 
top are used. They are often called domes. 


LIGHT-WEIGHT FLOOR CONSTRUCTION 603 


The floor consists of joists from 4 to 6 in. wide, spaced about 
26 in. on centers and connected by a thin concrete slab 2 to 3 in. 


' 
\ 
Ss 





| | 
aoe | l | | Lo 
os | 
jie | | aad, 
=e. | 3) Lee 2 
{ th wee [ 
all | | Jp set Se ee 0 a en BOs PO eke al | | asian 
ed 3 A) Se OO ae re 
mew) Uy We oy ol on Oe ee 
Nea O ier | jets, Tie i." 
cS et ela il a 2 
We er 
as ber " 7T | | [- 
ee sods = | | | 
atone | eee a nee ete i es SE See al [ oe 
“lieth aegeal E Darees, Leto 
| N 
See | | eee hel Vv i fee" | | hae ige 
cate rr | oo er ee Sa | 
| | S ] 
if) die, Seaeee x be | te 
a | | = ee v — oe a | | —e 
_— | | a Se ee eee eee | | 
Lois) 6 A Thy AN ror 
| ‘s S I | 
--—A bees ed ELISE 1 died RE eA gue The at Fl ow Lees 
ony ae Diwan Chaka Mice. to. 8 al ee r-- 
| ‘> | 
l 





ae 






Les €=-— 90" --- 


Section Through Joists 


Section Through Girder 


Fig. 194.—Typical Panel for Steel Tile Construction. (See p. 602.) 


thick, serving as a compression flange for the joists. The depth of 
joists below the topping varies from 4 to 14 in., depending upon the 
span and the load, 


604 BUILDING CONSTRUCTION 


Steel Tiles.—Two types of metal tiles are on the market: the 
removable tiles, which act only as a form and are removed after the 
concrete has hardened; and the permanent tiles, which are left 
permanently in the construction. Both types are of the general 
design shown in Fig. 195 and differ only in the strength of the sheet 
metal from which they are stamped. The removable tiles are made 
from sheet metal of 14 to 22 gage. The heavier tiles are preferable, 
as they last longer and keep their shape better after rehandling. 
For tiles left. permanently in the construction, sheet metal of 24 to 
26 gage is used. Permanent tiles must not be made too thin, or 
they will become distorted during construction. To stiffen the 


Holes about 
+ Diam. 







2 "depth of Corrugation 70 
. 






13- een a" \ 
Ff ~~ 37 abt. °EF3 * 
fe Red te Ok ak ak en 36” Covering Rad. = 
Width ; 
Pa Bae Se 


End View 





Jn a aa ; a aeceelts, 


Side View End View 
Tapered Steel Tile 


Frc. 195.—Details of Steel Tile. (See p. 604.) 


steel tiles and enable them to withstand the hard usage they receive 
during construction, they are provided across their tops and sides 
with corrugations formed by heavy dies during fabrication. 

The sides of the tiles are tapered and rounded at the junction of 
the slab and the joist. At the bottom, the tiles are provided with 
flanges one inch wide, which during construction rest on the forms 
and are lightly tacked thereto. Each row of tiles is closed at the 
ends by caps made of the same material and fitting the tiles. ‘The 
cap is shown in Fig. 195. 

Stock Dimensions of Steel Tiles. 

Standard Heights adopted by all manufacturers are 4, 6, 8, 10 
and 12 in. Some manufacturers also carry 14-in. tiles in stock. 


LIGHT-WEIGHT FLOOR CONSTRUCTION 605 


In other respects, the standards adopted by various manu- 
facturers differ to a great extent. 

Standard Width, measured at the bottom between the edges of 
the sides (not including the width of the one-inch flanges) varies 
from 20 in. to 25 in. This difference in width affects the spacing of 
the joists in the floor and should be taken into account when com- 
paring the strength of designs made by different firms. It is obvious 
that, with the same design for the joists, the floor will be stronger 
where the joists are spaced closer. For equal strength, the joists 
spaced farther apart should be thicker and have more reinforcement. 

In addition to tiles of standard widths, each manufacturer carries 
special tiles to be used in the last row near the edges of the panel, 
where the regular tile is too wide. 

Standard Lengths adopted by various manufacturers vary from 
2 ft. to 4 ft. 6 in. Some manufacturers use two standard lengths. 
In one instance, they are 30 and 36 in. of covering length (the actual 
lengths are 13 in. greater to allow for lapping). By combining the 
proper number of tiles of each length and by selecting proper lapping, 
forms for any length of joist may be obtained. 

Tapered Tiles——In addition to tiles of constant cross section, 
tapered tiles are manufactured. These are used at the ends of the 
joists where widening of the joists is required. The tapered tiles at 
one end are of the same width as the regular tiles, and at the other 
end the width is reduced by about 5 in. ‘This taper is obtained in a 
length of 30 in. Often the tiles are tapered in height also. 

The purpose of the taper is to increase the width of the concrete 
joist at the support, where additional shearing area is required. In 
continuous joists, the increased width also relieves the large com- 
pression stresses at the support. 

Adjustable Metal Tiles.—Another type of metal tiles is shown in 
Fig. 196, p. 606. The differences between these and the tiles just 
described are as follows: The adjustable tiles are made of smooth, 
heavy sheet steel, No. 12 gage, instead of the lighter corrugated 
material used in other tiles. They are wider and longer. Their 
width is 31 in., against 20 to 25 in. used for the other tiles. The 
length is from 6 to 10 ft. The depth of tiles is adjustable, as is 
evident from Fig. 196. It should be noticed that the metal tiles 
may be removed without removing the form under the joist. 

The metal forms are shown in place in Fig. 197. The pleasing 
appearance of the ceiling in the photograph, Fig. 198, is due to the 


606 BUILDING CONSTRUCTION 


use of smooth metal for the tiles and to the small number of joints. 
Both photographs were obtained from Mr. Wm. H. Gravell, engineer 
for the two jobs. | 

Method of Design.—The design of the floor system resolves 
itself into the design of the joists composing the slab and of the 
girders carrying the joists. 

For the purpose of design, the joists may be considered as small 
T-beams in which the topping serves as a flange. The following 
steps should be taken in designing: 

(1) Depth and width of the joist should be assumed for the pur- 
pose of determining the dead load. (2) Bending moments and 


7 #12 Smooth Sheet Steel 60" wide 








2 


zy 


~ 
~) 
Ss 
Le _~ Bottom of 
= Joist 
Lengths of 6; 8' & 10° 
hake 
ne lea” Removable ------.___ 
Support for 


Centers 


Wood Shoring and Centering 


Fic. 196.—Adjustable Metal Tile. (See p. 605.) 


shears should be computed. For continuous joists, use formulas on 
p. 278. (3) Shearing unit stresses should be determined. (4) 
Required amount of steel and the compression stresses in the center 
should be found by means of the T-beam formulas. (5) For con- 
tinuous joists, compression stresses at the support should be com- 
puted by means of formulas for rectangular beams with steel in top 
and bottom. 

Proper selection of the depth of the joist requires judgment. 
The depth must be sufficient not only to resist stresses but also to 
prevent excessive deflection. The remarks made in connection with 
hollow tile floors apply equally here. 

If shearing unit stresses are larger than allowed for concrete, 
either stirrups should be used or the width of the joist should be 


LIGHT-WEIGHT FLOOR CONSTRUCTION 607 





Fic. 198.—Ceiling in Penn Charter Gymnasium. (See p. 605.) 


608 BUILDING CONSTRUCTION 


increased by the use of tapered tiles. The bent bars in continuous 
joists are seldom available for diagonal tension reinforcement, 
because, to be effective as negative bending moment reinforcement, the 
bent portion must be some distance from the points of maximum shear. 
In continuous joists, compression stresses at the support must be 
provided for. To reduce compression stresses in concrete, either 
compression steel may be used or the width of the joists may be 
increased by the use of tapered tiles. In this connection, refer to 
p. 594, where the same ‘subject is treated for hollow tile floors. 
Points of inflection should be computed as explained on p. 595. 
Reinforcement for the joists should consist of bars of small 
diameter, else the bond stresses will be excessive. The bars should 
be placed at least one inch in the clear from the edge of the joist. 
Not more than two bars should be placed in a joist 5 in. wide. Bars 
should be held securely in place. In all respects, the design should 
comply with the requirements eiven for hollow tile joists on p. 594. 
Design of Topping.—The topping, i.e., the thin concrete slab 
between the joists, serves two purposes: first, to carry the load 
placed thereon to the supporting joist; and second, to act as com- 
pression flange for the joists. For all practical purposes, the topping 
‘s a concrete slab restrained at both supports (i.e., at joists), and 
carrying the load on a span equal to the distance between the edges 
of the joists plus the thickness of the topping. Since it is not prac- 
ticable to reinforce the thin slab for negative bending moment, to 
prevent cracking, the thickness of the slab should be such that the 
tensile stresses in the concrete, due to negative bending moment, 
will not exceed the safe value for plain concrete. (For 2 C00-lb. 
concrete this is 40 lb. per sq. in.) The minimum thickness of the 
topping should be 2 in., exclusive of the finish. For a width of core — 
of 20 in., this thickness is good for a live load up to 150 Ib. per sq. in. 
For larger live loads, the thickness of the topping should be increased. 
For larger spans, the safe load is proportionally smaller. In many 
designs, the bending stresses in the topping are disregarded, and the 
concrete is subjected to considerable tension. The same disregard 
is found in tables prepared by some manufacturers of steel tiles. 
For instance, in one table, a live load of 666 Ib. per sq. ft. is recom- 
mended for joist construction with a 9-in. topping, although this 
load would destroy the thin slab. 
The topping acts also as a flange for the joists, and its thickness 
must be sufficient to resist the compression stresses developed by 


LIGHT-WEIGHT FLOOR CONSTRUCTION 609 


the bending moments in the joists. For deep joists of long span, the 
thickness of topping may have to be increased to keep the compres- 
sion stresses within working limits. 

From the above, it is evident that the topping is a vital part of 
the construction, and that it is of importance to prevent its separa- 
tion from the joist. Therefore, shrinkage stresses and temperature 
stresses must be resisted by reinforcement placed at right angles to 
the direction of the joist. This may consist either of small bars or 
of wire mesh. ‘The cross-sectional area of the steel parallel to the 
joist should not be less than 0.2 per cent of the cross section of the 
topping for cases where steel tiles are left in the construction and the 
ceiling is plastered. When the under side of the topping is exposed, 
at least 0.3 per cent of temperature steel should be used. 

In competitive designs, a small amount of temperature steel is 
used, such as {-in. bars 12 to 18 in. on centers. This practice pro- 
duces structures having a much lower factor of safety than required 
in other types of reinforced concrete construction. The small 
saving thus obtained is out of proportion to the harm done to the 
strength of the floor. 

When granolithic finish is used, it should not be considered as a 
part of the topping. 

Concentrated Loads.—The steel tile floor is not well adapted 
for concentrated moving loads. A concentrated load applied on the 
thin slab would overstress the concrete in bending. Another dis- 
advantage is that each joist would have to be designed to carry the 
full concentrated load, as the thin topping is not capable of bringing 
the adjoining joists into action. For this reason, steel tile floors 
should not be used for concentrated moving loads. 

When the position of a fixed concentrated load is known, the joist 
under the concentration should be strengthened. For instance, 
under partitions, double joists are often used. When the concen- 
trated load comes between the joists, the topping should be made 
thick enough to carry the load and should have the proper amount of 
reinforcement. If the position of the concentrated loads cannot be 
fixed in advance, or if there is a likelihood of its being shifted, it is 
best to use other types of construction better adapted for concen- 
trated loads. 

Construction.—Since the slab is thin and the ribs narrow, par-. 
ticular care should be given to the pouring of the concrete and the 
placing of the steel. 


610 BUILDING CONSTRUCTION 


Topping Must be Poured Monolithic with the Joists Below.— 
The number of construction joints should be as small as possible. 
If it is necessary to place a construction joint at right angles to the 
joist, it should be placed in the center of the span. Construction 
joints parallel to the joist, when unavoidable, should be placed mid- 
way between the joists. 

To fill properly the spaces between the bars in the narrow joists, 
the concrete should be composed of coarse aggregate passing through 
3_in, mesh. Large stones may bridge in the forms and cause pockets. 
It is particularly important to force the concrete below the rein- 
forcing bars in the joists. Puddling during placing of concrete is 
advisable. 

The steel tiles must be removed with care, so as not to injure the 
thin topping and the edges of the joists. | 

Examples of Steel Tile Floor Construction—An example of the 
use of the steel tile floor construction is shown in the designs of the 
hospital building on page 807. Another example is the design of 
the building for Barnes Foundation, shown on p. 630. 




















Plaster on 
Rib Lath, 


Permanent” 


Lath with 
" Ri Metal Tile 


3 
: Rib. ~ 
































Before removal After Plaster 
of Forms Applied 


Fra. 199.—Method of Forming Flat Ceiling when Permanent Metal Tiles are 
Used. (See p. 610.) 


Flat Ceiling. —In construction with metal tile floor, a flat ceiling 
is obtained by attaching metal lath to the concrete and plastering 
the lath with three coats of plaster. 

When permanent tiles are used, metal lath is laid on the top of 
the wood forms before the metal tiles are placed. (See Fig. 199.) 
The tiles and the reinforcement are then placed, and the lath securely . 
wired to the reinforcement of the joist. The concrete, when poured, 
engages the meshes of the metal lath; but this bond should not be 
relied upon to carry the plaster and should not be considered as 
replacing proper wiring of the lath to the reinforcement. After 
the forms are removed, the ceiling is ready for plastering. 


COMBINATION OF STRUCTURAL STEEL AND CONCRETE 611 


The strength of lath for the ceiling should be sufficient to carry 
the plaster without undue deflection on a span equal to the spacing. 
The weight of the plaster, for ordinary three-coat work without 
ornamentation, amounts to 10 lb. per sq. ft. To avoid the use of 
furring channels, special types of lath with raised ribs are on the 
market. ‘They are sold under various trade names, such as Herring- 
bone Lath, Hy-Rib, Riplex, and others. Thir common property is 
that the expanded metal sheets are stiffened by rigid integral ribs 
running longitudinally with the sheets. The height of these ribs 
varies, depending upon the different uses to which they are put. 
For ceilings, the height of the ribs is usually 2 in. The thickness of 
material varies with the span between the joists. The gage of metal 
required for spans used in common practice is: for spans up to 24 in., 
No. 28 gage; for spans between 24 and 27 in., No. 26 gage; for spans 
up to 32 in., No. 24 gage. 

The rib of the metal lath should be placed against the support, 
and the mesh away from it. With this arrangement, the mesh offers 
a flat surface for plastering. The ribs are placed at right angles to 
the joists. Each rib should be anchored to the concrete. 

When removable tiles are used, the wire lath must be attached 
under the joists after the forms are removed. For this purpose, 
wire hangers, provided on one end with a hook, are installed in the 
concrete. After the metal tiles are in place, the wire hangers are 
pushed through special holes, which are provided for the purpose in 
the tiles, either on the top or at the sides, and placed so that the 
hook becomes inbedded in the concrete. After the metal tiles are 
removed, furring rods or channels, spaced proper distances apart, 
are suspended from the hangers and metal lath attached to them. 
With the rib lath described above, furring bars or channels may be 
dispensed with. 

If it is desired to have the ceiling some distance below the joists, 
suspended ceiling, as described on p. 613, should be used. 


COMBINATION OF STRUCTURAL STEEL AND CONCRETE 


A structural steel frame for beams, girders, and columns may be 
preferable to reinforced concrete for certain structures, such as city 
office buildings, simply for the reason that if the steel is fabricated 
in advance the buildings can be erected more rapidly although at 
higher cost. : 


612 BUILDING CONSTRUCTION 


In structural steel buildings, slabs between beams are usually 
puilt of concrete. The design of a panel with steel beams and girders 
is shown in Fig. 200. Concrete slabs are reinforced in the direction 
perpendicular to the beams. The main slab reinforcement should 
be run over the top of the beams and near the upper surface of the 
slab so as to make the slabs continuous. Transverse temperature 
steel is used, as recommended in connection with concrete slabs. 
Often, wire fabric or expanded metal are used for slab reinforcement. 













Bars’ 


for Slab 


Directions 


Main Reinforcing 











below Bottom Flange 
Fic. 200.—Typical Panel with Steel Beams and Girders and Concrete Slab. 
(See p. 612.) 


Note: Main Reinforcement shown schematically by outside bars only. For typical arrange- 
ment see p. 210. 


The cross section through slab and steel beams is shown in Fig. 
200a. In such construction, the steel beams and girders are designed 
separately as plain steel sections, as, due to lack of bond, no allowance 
can be made for strengthening effect of concrete. The slab is 
designed according to the principles of reinforced concrete. New 
York City, however, allows special empirical rules for design of 
cinder concrete slabs supported by steel beams. 

If the panel is square, or nearly so, the steel beams may be placed 
on four sides, as in Fig. 2006 above, and the slab is then reinforced 
by bars running in two directions, or by an adaptation of the Smulski 
System consisting of radials and cireles as shown in Fig. 128, p. 369. 


COMBINATION OF STRUCTURAL STEEL AND CONCRETE 613 


The slab between beams also may consist of two-way tile and con- 
crete floor as shown in Fig. 201, p. 613. 

Floors may consist of a combination of steel girders and rein- 
forced concrete beams and slabs, in which case proper seats in the 
steel frame are provided for the concrete beams. 

The slab between steel beams is sometimes constructed in the 
form of a concrete arch. If not reinforced, the beam should be con- 
nected with tie-rods, to resist any possible unbalanced thrust. For 
arches with curved upper surfaces, a fill of cinders or a very lean con- 
crete is used for leveling. 





Fre. 201.—Details of Two-way Hollow Tile Floor Slab and Structural Steel 
Column Fireproofing. (See p. 613.) 


SUSPENDED CEILING 


The term “Suspended Ceiling” is used when the plaster ceiling is 
more than 6 in. below the supporting floor construction. Its construc- 
tion differs from the plaster ceiling directly attached to the floor con- 
struction, such as shown in Fig.201, p.613. Suspended ceilings are used: 

(1) To provide flat ceiling in case of beam and girder or joist 
construction in which case it is placed below the deepest beam. 

(2) To provide ornamental ceiling. 

(3) To lower a part of the ceiling. 

(4) To provide air space below the roof construction for preven- 
tion of condensation. 


614 BUILDING CONSTRUCTION 


The suspended ceiling consists of plaster applied on metal lath 
which is suspended from the floor construction. The lath should 
be lapped both longitudinally and laterally so as to present a con- 
tinuous lathed surface. To get proper rigidity and strength the 
lath must be strengthened by furring as described below. All fur- 
ring and its supports and hangers must be strong enough to carry a 
load of at least 12 lb. per sq. ft. without undue deflection. The 
plaster will not crack for deflection up to sy of the span of the fur- 
ring. For ornamental ceilings, specially where false beams are used, 
the weight may exceed 12 Ib. per sq. ft., for which proper allowance 
should be made in design of the furring and hangers. 

The simplest design of suspended ceiling consists of lath, provided 
with longitudinal ribs or stiffeners, wired to furring channels, angles or 
flats placed on edge. Usually 3-in. furring channels spaced about 
2 ft. on centers give satisfactory results. Each rib or stiffener of 
the lath must be wired at each intersection with the furring channel 
using a 16 gage galvanized or annealed wire. The furring channels 
are suspended from the floor construction by 7 gage wire hangers 
spaced about 2 ft. on centers along the furring channel. The wire 
hangers on one end are imbedded securely in the concrete of the 
floor construction before the concrete 1s deposited. The free end is 
securely wound around the furring channel. | 

To insert the hangers, holes are drilled in the forms proper 
distances apart through which part of the wire is made to project 
the required length outside of the form. The end to be imbedded in 
concrete is preferably looped around the reinforcement of the floor 
construction. 

When the lath is not stiffened, furring rods 2 in. diameter spaced 
12 in. on centers should be placed across the furring channels and 
attached to them at each intersection. The lath is then wired to 
the furring bars. 

Where it is not desirable to use the large number of hangers 
required by the previously described method, following method may 
be used: The lath is attached to the 3-in. channels as in the previous 
case. These, however, instead of being suspended from the floor 
construction, are wired to from 13 to 2 in. channels spaced about 
8 ft. on centers and suspended from the floor construction by 4-in. 
hangers. The spacing of hangers along the channel is about 4 ft. 
on centers. | 

Where wire hangers are not provided ahead of time, holes are 


COMBINATION OF STRUCTURAL STEEL AND CONCRETE 615 


drilled in the concrete ceiling and expansion bolts are inserted. If 
the floor construction consists of hollow clay tiles, toggle bolts should 
be used to hold the hangers. Care should be taken that the load 
concentrated at each hanger is not large enough to break the 
tile. 

Suspended ceilings are shown in Fig. 202, p. 616. Simple ceiling 
as well as more complicated designs are shown. The construction 
of false beams is also shown. 

If more rigid construction is required, the wire hangers may be 
replaced by flat steel hangers which are ordinarily 1 in. wide and 
1 or 38; in. thick. The hangers are provided with holes for 3-in, 
diameter bolts to be used for securing them to the furring channels. 
Special design of the supporting members must be made when the 
suspended ceiling is unusual in any respect either on account of the 
weight of the ornamentation or because the spacing of the supporting 
members is large. Ordinary principles of design should then be 
employed. | 


INSERTS FOR ATTACHING SHAFTING AND SPRINKLERS 


When it is desired to suspend shafting or sprinkler systems from 
the ceiling it is best to make provision for it by imbedding in concrete 
special inserts spaced proper distances apart. _ 

Inserts may be of the threaded type where the bolts are screwed 
into the threads in the body of the inserts. Such inserts may be 
used only where no great accuracy in placing is required. To take 
care of any misplacement of the inserts during construction, special 
adjustable slotted inserts are used, in which the bolt may be moved 
several inches and thereby adjustment can be made. 

In cases where it is not possible to fix the position of the shafting 
ahead of time, continuous inserts are used. ‘These may be of any 
desired length and may even extend the full length of the building. 
With sufficient number of parallel rows of continuous inserts, shaft- 
ing may be placed in any desired location. 

The inserts must be attached firmly enough to the forms, so that 
no misplacement.can take place during concreting. The simplest 
method is by nailing them to the forms with two or three nails. To 
facilitate this, many inserts have special projecting lugs for the nails. 
To save the labor of cutting the projecting nails after the forms are 
stripped, special means of attaching are developed. In some inserts 
a tight fitting wood block is placed within the slot of the insert. This 


616 BUILDING CONSTRUCTION 














Furring 
Channel --y 


CaN VEZ 





Ceiling Suspended Below 
Beam from Concrete Slab 


Ceiling Suspended from 
Concrete Slab 


TUT TM PT OL SSL 
La SSS 


——— 










TAIT I So SI TITS UTTT ATA I LL PL ELL 
Pista crete eet oe 

Hanoy =; VRP ee ey || ER 

SDE Ra FAT WIT EY PRI PT aa) aes 


A 574. .: Pa | 
Mea) CL MONLN ec 





eS » 3" 
~Ixj§ Strap Iron 


False Beam and Cornice Suspended 
from Concrete Slab 


‘Yaulted Ceiling Suspended "Section A-A- 
from Concrete Slab 


Fic. 202.—Examples of Suspended Ceilings. (See p. 613.) 


COMBINATION OF STRUCTURAL STEEL AND CONCRETE 617 


block is provided with nails or screws which are fastened to the form. 
After the forms are stripped, the wood block is pulled out of the 
insert within the nail. 

There is considerable difficulty in attaching inserts to steel forms. 

An adjustable insert and a continuous insert are shown in Fig. 
203, below. 

When no provision is made for attaching the shafting, small 
holes may be drilled in the ceiling for expansion bolts. 

Where considerable load is to be supported through-bolts may be 
required extending all through the slab. In such case sleeves of 
proper diameter should be imbedded in the concrete to receive the 





Fria. 203.—Inserts Used in Concrete Construction. (See p. 617.) 


bolts. To prevent concrete flowing into the sleeves during con- 
ereting, they may be filled with sand. Very often through-bolts for 
attaching shafting are placed across beams. The hole for these 
through-bolts is also obtained by imbedding a sleeve in the beam. 

The inserts should be of such design that they would not pull out 
from the concrete under the load of the suspending shafting or 
sprinkler systen. In the inserts shown in Fig. 203 the loop in the 
individual insert extends some distance into the concrete while in 
the continuous insert the bolts extend into the concrete and serve 
as anchors. 

Protecting Exposed Corners.—Exposed corners of concrete col- 
umns or curbs should be protected by steel angles of proper dimen- 
sions. These angles must be anchored into the concrete by means 
of bolts. There are specially constructed curb bars on the market 
which are provided with projecting anchors integral with the bars. 


618 BUILDING CONSTRUCTION 


BASEMENT AND GROUND FLOOR SLABS 


This discussion applies to the basement floor slabs and to the 
first floor slabs in buildings without basement. 

The basement floor may rest directly on and be supported by the 
ground, in which case it may be built of plain concrete or reinforced 
with temperature steel only. If the ground is not firm enough to 
carry the slab load, reinforced concrete floor must be used. 

If the basement slab is subjected to upward water pressure, it 
must be designed for the upward pressure. 

Basement Floor Supported by the Ground.—The integrity of the 
slab resting upon the ground depends upon the firmness of the sub- 
grade and upon its uniformity. For this reason, the subgrade must 
be carefully prepared. All soft and spongy places must be removed, 
and the subgrade compacted by tamping. If the grade of the slab 
is above natural grade, the ground must be filled in with hard material. 
The use of soft material, trash, vegetable matter, and other perish- 
able material is prohibited. 

It is important to have the subgrade of uniform firmness, else the 
slab may settle unevenly and crack. If the subgrade is of sand or 
gravel, the basement slab may be laid directly upon it. A subgrade 
consisting of clay requires a subbase consisting of hard materials, 
such as gravel or clean, well-screened cinders. The subbase should 
be properly tamped and wetted before the concrete is laid. The 
wetting is helpful in compacting the subbase, but its main purpose 
is to prevent the absorption by the dry subgrade of water from the 
concrete. For this reason, the subbase or the subgrade, if a subbase 
is not used, should be wet at the time the concrete is placed upon its 

Plain Concrete Slab.—The basement slab may be built of plain 
concrete. To prevent cracking due to shrinkage and temperature 
changes, the floor should be divided into sections of not more than 
100 sq. ft. To prevent buckling and heaving of the slab in case of 
expansion, it is advisable to provide elastic joints at the wall and at 
the column. In slabs subjected to heavy truck traffic, the joints 
should be rounded to prevent their being clipped off. 

Basement Slab with Temperature Steel.—If it is desired to limit 
the number of joints the slab should be reinforced with temperature 
reinforcement placed in two directions at right angles. Reinforce- 
ment should also be used in all cases where it is not possible to get firm 
and uniform subgrade. The reinforcement then prevents cracks due to 


BASEMENT AND GROUND FLOOR SLABS 619 


bending, which occurs when a section of the subgrade yields more than 
the rest. 

The amount of reinforcement will depend upon the distance 
between expansion joints in the slab, upon the expected variation in 
temperature, and finally upon the load to be carried by the slab. 
The minimum reinforcement should consist of 2-in. round bars 12 in. 
on centers (or their equivalent). Wire mesh or expanded metal 
may be used to advantage. The reinforcement should be placed 
2 in. above the bottom of the slab, and should be properly lapped. 
The reinforcement must not cross the expansion joints; otherwise, 
the slab would not be free to move at the porns and their purpose 
would be frustrated. 

Thickness of Basement Slab.— Basement slab should be at least 
4 in. thick, including the wearing surface. 

Best results are obtained by using concrete of 1 : 2:4 mix for 
the base of the slab. The granolithic finish on the top should be of 
the same composition and should be laid in the same manner, as 
explained on p. 620 in connection with granolithic finish for rein- 
forced concrete work. 

In many instances, granolithic floor finish is not required, and the 
- basement slab is built for the full thickness of the same concrete. 
After the concrete is brought to the required grade, it is leveled off 
by a strike board and then floated and compacted with a wood float. 

Basement Floor Supported by Columns.—If it is not possible to 
get subgrade firm enough to support the load, the basement slab is 
supported on columns and designed in the same way as other sus- 
pended floors. The cost of the slab may be reduced by using 
intermediate short columns under the basement slab. For instance, 
in a 20 by 20 ft. span, intermediate columns placed in the center of 
the panel and halfway between the columns reduce the span of the 
slab to 10 ft. Of course, the use of intermediate short columns is 
economical only when foundation for them can be obtained without 
difficulty and at reasonable depth. Before deciding upon the type 
of construction, the saving due to reduction of span should be com- — 
pared with the cost of the additional column and foundation. Inter- 
mediate columns reduce the load on the main foundation. This 
should be considered when making comparison. 

Basement Slab Subjected to Water Pressure.—When the base- 
ment slab is subjected to upward water pressure, it should be rein- 
forced to withstand the difference between the upward water pres- 


620 BUILDING CONSTRUCTION 


sure at high water and the weight of the slab. The live load in the 
slab cannot be considered as contributing toward the balancing of the 
water pressure, becausé it is not always in place. 

The bending moments due to water pressure are of opposite sign 
to those produced by downward load. Thus, at the column, the 
bending moment is positive and requires steel near the bottom, while 
between columns it is negative and requires steel near the top. 

The construction of the basement slab to resist the upward pres- 
sure will depend upon conditions, such as the amount of pressure 
and the distance of the footings from the level of the slab. If the 
footings are some distance below the slab, they cannot be used to 
support the basement slab. The construction then must be tied to 
the columns. Flat slab construction will be cheapest if the spans 
permit, as it requires no formwork. Tor long spans, beams and 
girders may have to be used. 

If the columns are reasonably near the surface, the beams and 
girders may be attached to the footings. They must be secured by 
reinforcement against uplift. 

Sometimes it is necessary to design the basement slab for upward 
water pressure as well as for downward live load. The downward 
load to be used in design is the difference between the total live and - 
dead load and the water pressure at low water. ; 


FLOOR SURFACES 


In reinforced concrete buildings the surface finish of the floors 
depends upon conditions and also upon the use to which the building 
is subjected. Obviously, for example, different finish will be required 
for a hotel than for a warehouse. Also different portions of the same 
structure may require different finish. 


GRANOLITHIC FINISH 


A common and durable floor finish for concrete and reinforced 
concrete floors is the granolithic finish. This consists of specially 
proportioned and specially applied concrete. ° 

Two mixtures for granolithic finish are now in common use, namely: 

1. Mixture composed of one bag of Portland cement, and 2 cu. it. 
of fine aggregate such as very coarse sand. 

2. Mixture composed of one bag of Portland cement, 1 cu. ft. 
of fine aggregate such as coarse sand and 1 cu. ft. of still coarser aggre- 


FLOOR SURFACES 621 


gate consisting of clean, hard, crushed rock or pebbles. For still 
harder surface where appearance is of minor importance use one bag 
of cement, 4 cu. ft. coarse sand and 14 cu. ft. coarser aggregate. 
The size of the coarser aggregate should be such that it will pass 
when dry a screen having ?-in. openings and not more than 10 per 
cent will pass a screen having four meshes per lineal inch. 

The last named mixture gives harder floor finish and should be 
used where the floor is subjected to considerable wear, such as truck- 
ing. It is harder to travel and cannot be brought to so smooth a 
surface. It was designed originally by The Thompson and Lichtner 
Co. as a result of an extended series of tests with different composi- 
tions of material. 

Thickness of Granolithic Finish.—The thickness of granolithic 
finish depends upon conditions. It should not be less than { in. for 
integral finish, which is laid before the concrete of the structural slab 
is set, nor less than 1 in. for bonded finish, laid after the concrete slab 
has hardened. Where subjected to severe wear, an increase in thick- 
ness is advised. Thicknesses of 1 in. for integral and 13 in. for 
bonded are common practice. ' 

The “integral finish,” considered by itself is about one third 
cheaper than the “bonded finish,”’ due chiefly to the reduction in the 
amount of materials and saving in the cost of preparing base for 
“bonded finish.”? Another advantage of integral finish is, that there 
is no question of bond involved between the slab and the finish. 

In spite of these advantages the “bonded finish” is often preferred 
for the following reasons: The ‘integral finish”? must be applied 
before the concrete of the structural slab has set. Thus the finishing 
gang must work overtime in order to finish all the floor placed on 
that day. This adds to the cost especially where extra rates must 
be paid for overtime. Second, to avoid injury to the finish, the use 
of the floor for further construction work must be delayed. Third, 
when placing forms for upper floors it is not possible, even with proper 
precautions, to prevent entirely damage and scratching of the finish. 
Fourth, in inclement weather the finish is exposed to the action of 
the rain and frost, unless particular precautions are taken. F ifth, 
there is always danger that through carelessness or design, the 
“integral finish”? in part of the building at least will not be applied 
until the slab has hardened. All these items tend to reduce the 
economy of the “integral method.” | ? 

The “bonded finish” has these advantages. It is usually applied 


622 BUILDING CONSTRUCTION 


after the building is completed and closed in. It may be applied at 
any time and at any desired speed. The fresh finish does not need 
to be exposed to use until it has hardened sufficiently. No special 
protection for “bonded finish” is required as the building is closed in. 

The most serious objection to “bonded finish’’ is, that if not 
properly bonded to the base, the granolithic may peel off. However, 
if applied in the manner recommended in the Specifications for 
Granolithic Finish in Volume IJ, satisfactory results will be obtained. 
It must be emphasized, that granolithic finish should be applied by 
skillful men, as poor workmanship, especially when coupled with 
poor selection of materials, is usually the cause of trouble with 
granolithic finish. 

Also special emphasis should be placed on the requirement in the 
specification just referred to that all laitance and loose material 
should be removed, that the slab be thoroughly roughened and 
wetted, and finally that the surface be coated with neat cement 
paste just before placing the finish. It is more economical to remove 
the laitance and roughen the slab before the concrete becomes 
thoroughly hard. Two other factors are of vital importance, 
namely: Selection of an aggregate free from dust or very fine particles 
which rise to the surface with the cement so as to produce a “‘ dusty ” 
surface and the avoidance of a wet, sloppy mix which also prevents 
a thoroughly hard, dense surface. _ | 

Advantages of Granolithic Finish.—Floors with granolithic 
finish are sanitary. They may be easily cleaned by a hose. To 
facilitate this the floors are pitched to drain through scuppers. The 
pitch is usually } in. to the foot. 

Granolithic finish is durable. If properly built it will resist severe 
wear, such as trucking for an indefinite period. 

Disadvantages of Granolithic Finish.—Often granolithic finish is 
objected to by the users on account of larger conductivity of heat. 
This objection is gradually disappearing for floors above ground as 
it was largely psychological. If a building is heated uniformly the 
floor slab also maintains a uniform temperature. The objection is 
often removed by oiling or painting the floors, thus relieving the 
cold appearing color of concrete. 

Granolithic is less resilient than wood. Work places for women 
required to stand all day should be covered with some material such 
as linoleum or rubber. For offices, also, a covering of this character 
is generally employed. 7 


FLOOR SURFACES 623 


Coloring of Granolithic Finish—Granolithic finish may be 
colored by mixing with the dry material for the concrete, in the entire 
finish, a proper amount, to be determined by trial, of mineral pigments. 
The materials should be mixed dry and thorough mixing is essential. 
As the admixture of coloring matter reduces somewhat the strength 
of the concrete, it should be limited usually to 5 per cent by weight 
of the cement. For description of coloring materials, see Volume II. 

Painting of Floors——Concrete floors may be painted to any 
desired color. If proper paint is used and is properly applied, it is as 
permanent as on wood floors. Naturally it wears out under foot 
traffic. Floors should not be painted where they are subjected to 
heavy wear. In such case the integral coloring is preferable. 

Directions for Laying Granolithic Finish —Complete directions for 
laying granolithic finish is given in Volume II of this treatise. 


TERRAZO FLOOR FINISH 


“Terrazo” is another type of cement finish. It consists of 
irregular marble or granite chips imbedded in cement mortar and 
ground to a smooth even surface.. The finish is usually 12 to 2} in. 
thick and is applied in two layers. The first layer, or binder course, 
consists of 1:3 cement and sand mortar and is about 1 to 1% in. 
thick. After the binder course is placed, tamped and leveled off, 
the finishing course is applied. This is composed of cement and 
stone chips, in proportions about 1 to 3 which are spread evenly 
over the whole surface and rolled. After this, additional chips are 
spread over the surface and rolled in. 

The finishing course is allowed to harden for about 24 hours. 
Then the surface is ground by rotary rubbing machine. Small 
surfaces may be rubbed by hand. Carborundum brick is used for 
rubbing both by machine and by hand. 

The color of terrazo finish depends upon the color of the chips. By 
proper selection of the material various color effects may be obtained. 

- Terrazo finish may be applied on green concrete slab or after the 
concrete has hardened. In the latter case, the base should be pre- 
pared as for granolithic finish by roughening, cleaning and wetting 
of the surface. Before the first course is placed the surface should 
be wet and then covered with neat cement paste as described for 
granolithic finish. 

Method of placing terrazo finish is fully described in Volume II of 
this treatise. 


624 BUILDING CONSTRUCTION 


WOOD FLOOR FINISH 


Wood floor surfaces are often used for residences, apartment 
houses, offices and hospitals; less often for factories and seldom for 
warehouses. The top flooring usually consists of hardwood such as 
oak or maple. Beech and birch also are sometimes used especially 
in combination with maple, which they resemble in appearance and 
wearing qualities. 

Oak flooring is used in residences and more expensive buildings 
only. It may be either plain sawed or quartered. Plain sawed 
boards are obtained by sawing the whole log in parallel layers. 
Quartered oak is obtained by sawing the log parallel to the diagonal 
line in each quarter. 

Maple flooring, sometimes asl with other hardwood, is used 
in other types of buildings. 

Component Parts of Wood Flooring—Wood flooring consists of 
hardwood top boards, and wood sleepers and cinder concrete fill. 
Where heavy flooring is required, intermediate flooring, consisting of 
thin boards or heavy plank, is used. 

Dimensions of Hardwood Boards.—The boards may be either 3 
or 4 in. wide. The 3-in. boards give better results than the 4-in. 
width, but the advantages gained are not worth the extra expense. 
The thickness of boards depends upon the use of the building. 
Flooring nominally 1 in. thick is most commonly used. Sometimes, 
when the flooring is exposed to hard usage, the nominal thickness is 
increased to 14 in. The actual finished thickness is about 7 in. 
smaller than the nominal dimensions. 

The boards are usually tongue and groved and blind nailed. 

Wood Sleepers.—The wood sleepers are usually spruce and their 
nominal cross section is 2 by 4 in. They are dressed on top and 
bottom and leveled on both edges so that the width on the top is 
reduced to 3 in. To prevent the sleepers from rotting when im- 
bedded in concrete they should be creosoted or kyanized. | 

Intermediate Flooring—Intermediate Flooring usually consists 
of 1 by 6 or 1 by 8 (nominal dimensions) spruce boards, surfaced on 
both sides. 

Heavy planks, ordinarily used for intermediate floors, are spruce. 
They are commonly 2 by 6 in. or 2 by 8 in. nominal dimensions sur- 
faced on both sides. 

The life of the floor is increased if the boards are creosoted or 
kyanized. | 


FLOOR. SURFACES 625 


Description of Wood Flooring on Sleepers.—Wood flooring may 
be divided into one-layer flooring and two-layer flooring. 

One-layer Flooring.—One-layer flooring is satisfactory for light 
usage only. It consists of hardwood top nailed directly to 2 by 4 in. 
leveled sleepers. ‘These are spaced 16 in. on centers and are im- 
bedded in 2 in. thickness of cinder concrete. 

A ply of rosin-sized building paper is laid underneath the hard- 
wood flooring. 

Two-layer Flooring.—Two-layer flooring consists of hardwood 
top, intermediate creosoted softwood flooring placed diagonally and 
2 by 4 in. leveled sleepers imbedded in 2 in. thickness of cinder 
concrete. <A ply of rosin-sized building paper is laid between the 
hardwood and the intermediate flooring. The sleepers are spaced 
16 in. on centers. 

A modification of the two-layer flooring consists of hardwood top 
placed diagonally 2 in. creosoted planks, which are secured to 2 by 
4 in. leveled sleepers imbedded in 2 in. thickness of cinder concrete. 
The heavy plank permits an increase of the spacing of sleepers to 
20 in. This type of flooring may be used for heavy trucking. 

Sleepers Imbedded in Concrete of the Slab.—To reduce the cost 
of the flooring, the sleepers are sometimes imbedded in the concrete 
of the slab. This method, although it does away with the cinder 
concrete, and thereby reduces the dead load and the cost is not satis- 
factory for the following reasons: There is danger of injuring the 
concrete slab while imbedding the sleepers in partly set concrete. It 
is difficult to keep the screeds at proper level. The floor is apt to 
have a hollow sound. 

Laying Hardwood Floor on Sleepers.—The hardwood floor may 
be laid as follows: The top of the supporting slab should be lower than 
the finished floor by the thickness of the screeds and flooring plus 3 in. 

Sleepers are laid first. They are held a proper distance apart by 
temporary 1’ x 6’ boards placed on edge at right angles to the 
sleepers, and about 4 ft. apart. After the sleepers are leveled 
and wedged up in low places, the space between them is filled with 
cinder concrete mixed in proportion 1:3:5. This should be care- 
fully packed under and around the sleepers. To prevent sleepers 
from becoming loose in case the concrete should shrink away, heavy 
nails are driven into them sideways. These become imbedded in 
concrete and assist in keeping the sleepers in position. The cinder 
concrete should extend to within 2? in. from the top of the sleepers. 


626 BUILDING CONSTRUCTION 


After the cinder concrete has hardened, the sleepers are leveled and 
any irregularity removed. The flooring should not be placed until 
the concrete has dried sufficiently. 

Hardwood Floor on Tar Base.—This flooring consists of hard- 
wood top resting on plank under flooring, which is imbedded in hot 
“mixture of tar and sand. To prevent warping of the planks such 
flooring is often built with a diagonal intermediate floor placed be- 
tween the planks and the hardwood top. 

The thickness and quality of the hardwood top and of the inter- 
mediate flooring are the same as described on page 624 in connection 
with other types of hardwood flooring. The planks for the under- 
flooring are of spruce, 1% in. thick, surfaced on both sides. 

The layer of tar and sand is usually at least 1 in. thick. The 
planks are laid on the soft mixture about 4 in. apart and settled by 
hammering until stable bearing is obtained. After the planks are 
leveled they are edge-nailed together with 12d. common nails spaced 
12 in. apart. 

Wood Block Floors.—Wood block floors consist of blocks from 
2 to 3 in. thick laid in coal tar pitch on the concrete base. The 
blocks are made of natural redwood or of creosoted leaf and hard pine. 

Because of its wearing qualities the wood block floor is often used 
for special trucking aisles or spaces subject to heavy trucking in 
connection with floors elsewhere finished otherwise. The advantages 
of the wood block floor are exceptional durability under severe 
trucking and wearing conditions, resiliency, dustlessness, noiseless- 
ness and ease of cleaning. It is less slippery, either wet or dry, than 
most other surfaces. 

The following method of placing wood blocks is recommended 
by Mr. MacMillan.° 

The top of the concrete foundation shall be brought to a true, 
smooth and even finish exactly the depth below the finished floor level 
corresponding to the depth of the block to be used. This shall be 
accomplished by applying a mortar finish with a long handle wooden 
float or similar device. Care shall be taken to see that there are no 
projections in the concrete that will form an uneven bearing for the 
blocks. 

After the concrete has thoroughly dried out, it is swept clean and 
given a thin, even coating of coal tar pitch, not exceeding g of an 


5 Factory Floor Surfaces, by A. B. MacMillan, Chief Engineer Aberthaw 
Company. 


FLOOR SURFACES 627 


inch in thickness. The coating shall be allowed to harden before 
laying the blocks and shall not be applied over 20 ft. in advance of 
the block laying. 

Upon the base, as above prepared, the blocks shall be laid tightly 
together, with the grain vertical; the courses of the blocks shall be 
kept straight and parallel, starting from one side of the building and 
carried through to the other side; all joints shall be broken by a lap 
of at least 2 in. In truckways, and whenever possible, the blocks 
shall be laid with their length at right angles with the line of traffic. 

After every four rows of blocks shall have been laid in place, a 
piece of 2 X 4 in. planking shall be laid along the outside edge of the 
blocks and the courses driven together as tightly as possible. The 
blocks in each separate row shall also be tightened lengthwise just 
before the filler is applied, by forcing the blocks together from the 
ends, with a lever, pick or other instrument. 

Against the walls on all sides of the floor, as well as around all 
columns and other obstructions, a bituminous expansion joint 1 in. 
in width shall be formed by first laying a wood strip of that width, 
and after removal filling the space to within an inch of the top with 
coal tar pitch. © 

After the blocks have been laid in place and brought to as true 
and level a surface as possible, the joints between the blocks shall be 
filled with coal tar pitch (or equal material) applied at a temperature 
of not less than 350° F. The filler shall be applied by flushing over 
the surface of the floor, using a rubber-edged squeegee to force it 
into the joints. Care must be taken to see that the filler penetrates 
the full depth of the blocks and that the joints are completely filled 
at the time of application. With proper care, the floor space will be 
almost free from filler. Dry, sharp sand shall then be swept over 
the floor completely covering the blocks, and shall be left on the 
floor until the blocks are well set, if possible for a period of three 
weeks. ‘Traffic will, during that time, wear off the thin film of filler 
left on the surface. If any superfluous filler remains after the sand 
has been swept off and it proves obnoxious, steps may be taken to 
remove it with any one of the several approved methods. 


COMPARISON OF FLOOR SURFACES IN CONCRETE BUILDING 


A comparison of cost and durability of the various flooring 
discussed in the previous pages may be had from the table on page 
628, taken from Mr. A. B. MacMillan’s “Factory Floor Surfaces.” 


BUILDING CONSTRUCTION 


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CHAPTER XI 
WALL-BEARING CONSTRUCTION 


In buildings not exceeding three stories in height, the walls may 
be used to support the reinforced concrete floor construction. The 
construction is then called wall-bearing construction to distinguish 
it from the skeleton construction previously described. 

The advantage of this type is that the bearing walls dispense 
with the columns, thus reducing the cost. It has the disadvantage, 
however, that during construction the speed of erection of floors 
depends upon the speed in erection of supporting walls. The two 
parts of the job are usually of different materials and handled by 
different trades. It often happens that delay in erection of the wall 
holds up the whole construction. This cannot happen in skeleton 
construction, where the erection of the complete skeleton is inde- 
pendent of other parts of the job. 

The bearing walls are most often of brick, but they may be of 
concrete blocks, terra cotta blocks, or plain or reinforced concrete. 

This type of construction is often used in schoolhouses and hospi- 
tals and also in private dwellings. Fig. 204, p. 630, shows a plan of a 
wall-bearing job. 

The building illustrated is one of a group of buildings for the 
Barnes Foundation designed by Paul P. Cret, Architect, and William 
H. Gravell, Engineer. It consists of basement, two floors and roof. 
The floor construction is of the adjustable joist type described on 
page 605. It is supported by outside and inside brick walls. The 
details of the floor construction are clearly shown in the figure. 
Attention is called to the special type of stair construction in which 
the stair load is carried by beams placed at the edges of the stair slab. 
They are supported on one end by a floor beam and on the other end 
by brickwork. The stair slab is designed as spanning between the 
edge beams. 

Brickwork.—To serve as bearing wall, the brickwork must be 
made strong enough to carry the superimposed loads with a proper 

629 


630 WALL-BEARING CONSTRUCT ION 


factor of safety. The bricks must be hard and must have a crush- 
ing strength of single brick on the flat of at least 2 300 lb. per sq. in. 















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Typical Section Pe \Brich Bearing Wall~x, ——e=—e—! 


Frame. Plan of Second Floor 


Fic. 204.—Floor Plan and Details of Barnes Foundation Building. (See p. 629.) 
Paul P. Cret, Architect. William H. Gravell, Engineer. 


They should be laid in Portland cement mortar. The brickwork 
should be properly bonded. It is generally required that one course 
of headers must follow every five courses of stretchers. Face brick, 


REQUIRED THICKNESS OF BRICK WALLS .- 631 


if properly bonded to the rest of the brickwork by means of headers, 
may be considered as part of the thickness of the wall. If the face 
brick is tied to the wall by metal ties only, it must not be considered 
as effective in carrying the load. 

Stresses in Brickwork.—The allowable stresses in brickwork 
depend upon the hardness of the brick and also upon the strength of 
the mortar. For supporting concrete construction, Portland cement 
mortar mixed in proportions as rich as one part cement to three 
parts sand (1 : 3) should be used for bearing walls. 

The following unit stresses may be used: 


Allowable 
Stresses, 
Ib. per sq. in. 
Hard paving brick material (crushing strength of single bricks 
PU meer Meee. ee ltt. ok ngs ca ewe ee chee es 350 Ib. 
rere UM eee nee, ie ac. Ue OLR AA, 250 lb. 
Hard common brick (crushing strength of single bricks 2300 lb. per 
Sopa meee eee Nt! Sd oi is Payee! vale Aa ke 2 200 lb. 


Bearing Stresses on Brickwork.— When the available area of the 
wall is not less than twice as large as the area of application of a con- 
centrated load, the allowable bearing stresses produced by the load 
may be equal to 14 times the stresses recommended above. For 
smaller ratios between the area of wall and the area of application 
of the load, the allowable bearing stress may be reduced propor- — 
tionally. 

Required Thickness of Brick Walls.—The minimum thickness of 
the brickwork for outside walls of industrial buildings, warehouses, 
hotels, office buildings, schools, and hospitals should be as given in 
the table below: 


Base- | 1st 2nd 3rd 4th 5th 6th 











Height of Building ment | Floor | Floor | Floor | Floor | Floor | Floor 
Re sey co ben oi © 12 12 
PeOMILDTICS ee ee 16 12* 2 
Brose BLOTIES is. Su. ess oe. 16 16 12* 12 
Me tenes oii ce fe. el 20 20 16 16 12 
BEPPOreL OPTICS, fhe) ia ws bis koe 24 20 20 16 16 16 
SO i ares 24 20 20 20 16 16 16 


* For schoolhouses use 16-in. wall. 


632 WALL-BEARING CONSTRUCTION 


Limitations—The above table is valid for clear story heights 
up to 18 ft. For story heights from 18 ft. to 24 ft., 16-in. wall 
should be used instead of 12-in. For story heights from 24 ft. to 
30 ft., 20-in. wall should be a minimum. No changes are required 
‘n other items of the table. For walls 50 ft. long and under, the 
thickness of the wall may be decreased by 4 in. the minimum thick- 
ness, however, should remain 12 in. The length of the wall should 
be considered either as the total length or as the length between 
cross walls. | 

Walls used as enclosures for staircases, elevator shafts, or other 
shafts, when their length does not exceed 25 ft., may be built 12 in. 
thick for upper 50 ft. and 16 in. thick for the remainder of the height 
of the building. 

The thickness of the wall given in the table, or as modified by 
the above, should be considered as a minimum and should be increased 
if the stresses on brickwork exceed the allowable working stresses. 

Pilasters.—For heavy concentrated loads due to girders or 
trusses, pilasters, buttresses, or piers should be provided. The 
dimensions of these should_be such that’ the stresses on the brickwork 
are kept within working limits. In computing stresses, all dead load 
plus the proportion of live load recommended on p. 453 should be 
used. \ 

The wall between pilasters may be reduced in thickness by one- 
half of the difference between the thickness of pilaster and the 
required thickness of wall without pilasters. The reduced thickness 
of the wall should not be less than 12 in. for walls less than 30 ft. 
high and 16 in. for walls over 30 ft. 

When the thickness of the wall is reduced on account of the use 
of pilasters, their width must not be less than one-tenth of their 
spacing measured center to center. 

Interior Bearing Walls.—Interior bearing of brick walls used in 
concrete construction should not be less than 12 in. thick nor less than 
one-fifteenth of the clear story height. The thickness must be increased 
when required by stresses. In computing stresses in brickwork, the 
reduction in live load recommended on p. 453 may be used. 

Construction at the Wall.—When the ends of concrete beams rest 
on a’ brick wall, the bearing stresses should be investigated. If 
they are larger than the allowable stresses, as given on p. 631, the 
bearing area should be increased. This may be accomplished in 
several ways, as described below. 


CONSTRUCTION AT THE WALL 633 


Steel bearing plates or steel shapes may be used to distribute the 
load on brickwork; the width of the beam may be enlarged; or, if 
this is not sufficient, projections may be provided on both sides of the 
beam. Such projections should be of the same depth as the beam, — 
to facilitate the laying of brickwork above the floor. 

The thickness of the bearing plates and the dimensions of the 
concrete projections may be determined by considering the part out- 
side the beam as loaded by the upward reaction of the brickwork 
and supported at the beam. It may be necessary to reinforce the 
concrete projections by bars placed near the bottom, parallel to the 
wall. 

A continuous distributing beam running along the wall may be 
used, especially in case of joist construction. This should preferably 
be of the same depth as the beams or joists. (See Fig. 204, p. 630.) 

A continuous distributing beam may be considered, for design- 
ing purposes, as a continuous beam loaded by the upward reactions, 
and supported by beams or joists. The reinforcement will therefore 
be required near the bottom at the supported beam, and near the top 
between the beams. Usually, however, continuous bars are placed 
near both top and bottom 

Where the distributing beam spans an opening, its action is 
reversed and it serves to carry the loading. Its dimensions will be 
governed, in this case, by the load and the span. 

If possible, the distributing beam should be of the same width 
as the wall, for the sake of simplifying the construction. When it is 
not desirable to expose the concrete, the width of the beam is made 
41 in. smaller than the wall, allowing for brick veneer. 

The advantage of the distributing beam is that it ties the con- 
struction and in a measure prevents unequal settlement of the brick- 
work. It is always used in joist construction. The brick is stopped 
at the bottom of the joists, and for the next story is started on the 
top of the concrete. This gives a much neater job than if an attempt 
were made to brick in the spaces between the joists below the slab. 
The distributing beam should be reinforced with at least four j-in. 
round bars, two near the top and two near the bottom. Of course, 
special provision must be made over windows and doors, depending 
upon the span of the opening and the load to be carried. 

Provision for Negative Bending Moment at the Wall.—The brick- 
work placed above the concrete, especially when distributing beams 
are used, prevents free movement of the concrete construction. 


634 WALL-BEARING CONSTRUCTION 


The amount of restraint so produced is problematic and depends 
upon the amount of brickwork and the load which it carries. It is 

larger in the lower stories and decreases toward the top. Therefore, 
it is not advisable to rely on it in computing the beams. ‘To prevent 
cracks, however, it is advisable to use a certain amount of reinforce- 
ment near the top of the beams and also along the distributing beams. 
One-half of the steel used at the first interior support will be found 
sufficient in ordinary cases of brick bearing jobs. This amount of 
steel corresponds to a bending moment for uniform loading equal to 

wi? 


Ye= 30° (See also p. 278.) 


CONCRETE WALLS 


Reinforced Concrete Bearing Walls.—Reinforced concrete may 
be used for bearing walls. Their thickness may be governed by the 
stresses, as in concrete columns. If built monolithic with the floor 
system, they may be exposed to bending and should be reinforced 
accordingly. ‘Temperature steel should be used. Special rein- 
forcement is required at the corners of the building to prevent the 
separation of the walls. Where a cross wall and a longitudinal wall 
are poured together, the horizontal bars of the cross wall should be 
extended into the longitudinal wall and hooked at the ends to pre-. 
vent a crack. If it is not desired to have a connection between the 
walls, they should be poured separately so that they may expand, 
contract, and settle independently. A recess should be provided in 
the longitudinal wall; a few dowels may also be used to prevent the 
opening of cracks between the walls, but their effect should not be 
sufficient to make the walls cooperate. 

Special reinforcement should be used above door and window 
openings. For wide openings, the necessary amount of reinforce- 
ment is found by considering the lintel as a beam. The reinforce- 
ment from the lintel should be run a sufficient distance into the wall 
to prevent a temperature and shrinkage crack at the junction. 

The thickness of the concrete bearing wall may be determined by 
stresses, the ratio of slenderness being taken into account. Some 
codes specify that the thickness of the conerete wall shall be three- 
fourths of the thickness required for brick wall. 

Reinforced Concrete Enclosing Walls.—The enclosing walls are 
sometimes built of reinforced concrete. When supported on the 


CONCRETE WALLS 635 


framework, they do not need to be more than 6 in. thick, this thick- 
ness being sufficient to make the wall weather-proof. The walls 
should be anchored at all floors and reinforced with bars placed near 
their outside and inside faces; they should be made strong enough 
to resist a pressure of 30 lb. per sq. ft. applied either from the out- 
side or from the inside. This pressure represents the wind pressure 
and any accidental pressure that may be exerted from the inside. 
Temperature steel is usually sufficient to resist any stresses due to 
such causes. : 

Recesses for the walls should be provided in the columns; also 
bonds or dowels (1.e., short steel bars) above the floor beam and below 
the spandrel should be used. If walls are used above and below a 
beam, the same bars may be used for bonds above the floor and below 
the spandrel. The bonds may consist of 2-in. round bars spaced 
24 in. on centers. These should be placed so that they come in the 
center of the wall. 

Temperature Reinforcement for Enclosing Concrete Walls.— 
When concrete walls consist of panels between columns which are 
built separately, the amount of reinforcement given below may be 
used. 





Spacing of ? in. Round Bars 


Thickness of at Each Face 


Wall 
Horizontal Bars Vertical Bars 
6 in. 18 in. 24 in. 
8 in. 14 in. 24 in. 
10 in. pee Line 24 in. 
12 in. 9 in. 24 in. 





If walls are monolithic with the columns, at least one-quarter of 
one per cent of horizontal reinforcement is required to distribute the 
cracks properly. These should lap at the columns. 

Reinforced Concrete Fireproof Partitions.—Reinforced concrete may 
be used for partitions, especially around staircases. The minimum 
thickness allowed by most building codes for reinforced concrete 
partitions is 33 in. Such partitions are considered fireproof and may 
be used for enclosing staircases, shafts, and elevator wells. The 


636 WALL-BEARING CONSTRUCTION 


authors recommend in preference a thickness of at least 4 in., to give 
better space for pouring concrete around the reinforcement, with steel 
bars 2 in. round placed 12 in. on centers horizontally and 24 in. on 
centers vertically. The steel should be placed in the center of the 
wall. Special corner bars should be provided at the corners of the 
shaft or else the bars from one wall should be bent at right angles 
and extended into the other wall. The walls should be poured after 
the floor construction 1s completed. Bonds consisting of short bars 
spaced 2 ft. on centers should be provided above the floor and below 
the ceiling to tie the wall to the floor. Bars may be inserted into the 
pour hole. Pour holes spaced about 2 ft. on centers should be left in 
the slab above the wall to pour the concrete, although this is not 
necessary if the wall is under a beam, in which case holes may be left 
in the top of the wall forms and spouts arranged to receive the con- 
crete. ! 

Special reinforcement consisting of 4 or % bars should be provided 
at the top and sides of door and window openings. Short diagonal 
pars at the corners may also be used advantageously to prevent 
cracks. 


CHAPTER XII 
BASEMENT WALLS 


Basement walls, that is, walls below grade, in reinforced con- 
crete buildings or in steel buildings, may be thinner when made of 
reinforced concrete than when made of plain concrete. The prin- 
cipal reinforcement is supplied to resist stresses due to earth pressure. 
Some additional temperature and shrinkage reinforcement is needed, 
the amount depending upon the degree to which elimination of 
shrinkage cracks is desired, and also upon the design. More tem- 
perature steel is required for basement walls built monolithic with 
the column than for walls built separately between previously 
poured columns. 

Method of Construction.—The basement wall may be built either 
monolithic with the basement columns or separately from them. 
The latter method reduces the cost of wall forms. When the wall 
is built separately from the columns, proper recesses along the column 
should be provided. These recesses are formed by nailing, on the 
inside of the column form, a strip of wood of proper width and thick- 
ness. Dowels may be used to tie the wall and column. The joint 
between the column and the wall may be kept waterproof by insert- 
ing in the recess proper plastic waterproofing material. 

Thickness of Wall.—To be impervious to ground water, the 
basement wall should be at least 8 in. thick. Ordinarily, a 12-in. 
wall is used. This dimension may have to be increased if larger 
depth is required to resist earth pressure. 

In ornamental structures, the thickness of the masonry in ‘the 
first floor is sometimes greater than the required thickness of base- 
ment wall. If the difference is small, the greater thickness is adopted 
for the basement wall. If there is a large difference between the 
thickness of the masonry above and the basement wall, it may be 
economical to provide the basement wall at the top with a con- 
tinuous bracket to carry the masonry. (See Fig. 265, p. 754, and 
Fig. 290, p. 808.) 

637 


638 BASEMENT WALLS 


Earth Pressure.—For a basement wall below grade, the magnitude 
of earth pressure is determined in the same way as for retaining walls. 
The unit pressure will depend upon the character of the soil. Under 
ordinary conditions, earth pressure equal to the pressure of a fluid 
weighing 30 lb. per cubic foot 1s considered ample. 

When the wall abuts on a roadway or a railroad track, the moving 
load produces an additional pressure on the wall. ‘This can be taken 
care of by figuring a surcharge, as explained in connection with 
retaining walls (p. 838). For streets, a surcharge equal to 3 ft. of 
earth may be used. For railroad track, the surcharge should be 
from 6 to 10 ft. depending upon the distance from the track and the 
weight of the locomotive to be operated. 

Without surcharge, the pressure on basement wall varies accord- 
ing to a triangle. The pressure is zero at the ground level and a 
maximum at the level of the basement floor. There it equals the 
unit pressure multiplied by the depth in feet below the ground level. 
If the basement wall is 10 ft. deep and the unit pressure is 30 lb. per 
sq. ft., then the maximum pressure 10 ft. below ground is 10 X 30 
= 300 lb. per sq. ft. } 

The effect of the surcharge is the same as if the ground level 
were raised by the amount of the surcharge. Thus, with 3 ft. sur- 
charge and a unit pressure of 20 lb. per sq. ft., the pressure at the 
ground level is not zero, but 3 xX 30 = 90 Ib. Ten feet below 
ground the pressure would be (10 + 3) X 30 = 390 lb. per sq. ft., 
‘nstead of 300 Ib. for wall without surcharge. ‘The pressure, instead 
of being represented by a triangle as in the case without surcharge, 
is a trapezoid. 

Further explanation may be found in the chapter on Retaining 
Walls. 

Design of Basement Wall.—The direction of reinforcement to 
resist the earth pressure depends upon the manner in which the wall 
is supported. It may be considered as.a slab supported on the top 
and bottom, that is, on the first floor level by the floor construction 
-and on the basement floor level by the resistance of the basement 

floor and the passive resistance of the earth below the basement slab. 
Another method is to consider the wall as a slab, supported by the 
columns and spanning horizontally between them. The third 
method is to reinforce the wall in two directions. 

Wall Supported on Top and Bottom.—The first method, assuming 
the wall as a slab supported at top and bottom, is possible when there 


WALL SUPPORTED ON TOP AND BOTTOM 639 


are no wide openings in the basement wall. At wide openings there 
is no connection between the wall and the top slab; hence, there is 
no support for the wall at the top. This method is economical when 
the spacing of the columns is more than 40 per cent larger than the 
depth of the wall. Its use is also advisable when the basement 
columns are slender and would have to be materially strengthened 
if any earth pressure were transferred to them. The main rein- 
forcement is placed vertically, near the inside face of the wall. 
Some horizontal steel is needed to prevent cracking due to tempera- 
ture and shrinkage. Under ordinary conditions, this would consist, 
for a 12-in. wall, of 2-in. round bars, 12 in. on center. Other thick- 
nesses Of wall may be reinforced proportionally. For a watertight 
wall, at least 0.3 of one per cent should be used. The temperature 
steel should be placed near the inside face. To prevent cracking of 
the wall at the column, due to bending moment produced by earth 
pressure, short horizontal bars should be placed at the column near 
the outside face. 

On the top of the wall, the earth pressure is transmitted and 
resisted by the floor construction. At the bottom, reliable support 
for the slab must be provided. When the basement slab is at least 
6 in. thick and extends over the full width of the building, the base- 
ment wall may be considered as supported at the bottom by the 
basement slab. Ordinarily, it is not advisable to rely on the base- 
ment slab alone, because very often the earth behind the wall is 
filled in and earth pressure is exerted against it before the basement 
slab is poured. To avoid trouble due to this, the wall should be 
carried down into solid ground below the basement slab for a distance 
of at least 9 in. The passive earth pressure then resists the reaction 
of the wall. 

When the ground below the basement slab is not firm, it is 
not possible to rely on passive resistance of the ground for the 
horizontal support of the wall at the bottom. Unless the floor 
slab is designed to take the pressure from the wall there should 
be a beam extending from column to column and reinforced to 
transmit the load from the wall to the columns. Some designers 
provide a projection on the wall, on which the future basement 
slab will rest. If the slab is not reinforced and it rests on yield- 
ing ground, the projection is objectionable, because the slab after 
settling is likely to crack along the edges where it is held by the 
wall. 


640 BASEMENT WALLS 


Bending Moments Due to Earth Pressure.—When the slab is 
supported on top and bottom, the bending moment and reaction may | 
be found from the following formulas: 

Triangular Variation of Pressure-—When the earth pressure 
varies as a triangle, 

Let w = maximum unit earth pressure, lb. per sq. ft.; 

h = height of wall, ft.; 
R, = reaction at the top; 
Rz = reaction at the bottom; 
M = maximum bending moment; 
x1 = distance of point of maximum bending moment from top. 


Then 
1 wh 
Ry — 2 Lae . . . ° ° . ° e ° e . (1) 
2 wh. 
Roan La oss hg 
M = 0.064wh? ft.-lb. . 2 on 
= 0.7 7wh? inlb» 4.) soe ee 
Maximum moment acts at a distance below the top equal to 
A 
= ah = 0. 58h. f “ A : A 2 c Cs 5 


Trapezoid Variation.—When the earth pressure varies as a trape- 
zoid, in addition to the above notation, 
Let w, = unit earth pressure on top, lb. per sq. ft.; 
We = unit earth pressure on bottom, lb. per sq. ft. 





R, 2, 
6 

Ry = Mit Aes, _ , ORS be Seine a 

Minax = Se ft.-lb. (approx.). . . . (8) 

= 0.75 (w; + we2)h? in.-lb. (approx.) . . (9) 


The formulas for reactions are exact. The formula for bending 
moment is approximate and is based on the assumption that the 
total load is uniformly distributed. The formula for exact bending 
moment is complicated. 


WALL SLAB WITHOUT SURCHARGE 641 


The distance between the point of maximum bending moment and 
the top may be obtained from the following formula: 


1 pas 


oe 1+V2(2+o(¢—1)+1]h,. . (10) 


W2 
where c = —. 
Wi 


When this value is found, actual maximum bending moment may 
be computed by ordinary rules of statics. Ordinarily, the approxi- 
mate Formula (8), p. 640, gives accurate enough results. 


Example 1.—Design basement wall for a building in which wall columns 
3 ft. wide are spaced 20 ft. on centers. The distance from top of basement slab 
to top of first floor is 10 ft. and the top of the ground is 4 in. below the first floor. 
On one side of the building runs a railroad track producing an earth pressure 
equivalent to a surcharge of 6 ft. 


Solutton.—The wall, as shown in Fig. 205, p. 642, is supported at top and 
bottom. The theoretical support of the wall, at the top, may be assumed to be 
in the center of the slab, and at the bottom, 6 in. below the basement slab. The 
theoretical height of wall is h = 10 ft. 8 in. 

Wall Slab without Surcharge.—Taking the unit pressure at 30 lb. per sq. ft., 
the maximum pressure at the bottom is w = 30 X 10.67 = 320 Ib. per sq. ft. 
The pressure is represented by a triangle, for which the maximum bending 
moment, from formula M = 0.77 wh? (see p. 640). 


‘M =0.77 X 320 X 10.66? = 28 000 in.-lb. per lin. ft. 


Assume 12 in. for thickness of wall; then d = 10.75 in. and the required 


M 
A; =---- 
amount of steel, from As jaf, 
28 000 : 
As = (See = 0.19 sq. in. per lin. ft. 


0.875 X 10.75 XK 16 000 


1_in. round bars, 12 in. on centers, give an area of steel equal to 0.196 sq. in. 


per lin. ft. 


0.19 ; : : 
The ratio of steel to concrete is p = ORT 0.0015. Since this ratio 


is smaller than 0.077, the compression stress in the concrete is below the allow- 
able value, and the thickness of wall is sufficient. 
: 10.66 
The pressure transmitted to the ground at the bottom is 3 X 320 X re: 
1140 Ib. per lin. ft. Since the wall is imbedded 12 in. in the ground below the 
basement floor, the horizontal unit pressure on the ground is 1140 lb. per sq. ft. 
This is satisfactory. 


642 BASEMENT WALLS 


| 


I 
















i] 

Noe 
el | 
SS | | 
Sia | ' 
Sic 27801 | | 
lo per sq. ft. ‘Y | 
eat pro 
18 BES 

: a 


>| 8" 






—ec (ew ee we -—— — 


wo 


10 Ft. 8*- - 







-e===- 10 Ft. 0 


s{--7—-~--~-Span Length 


BOD Dot ae er With Surcharge 


oe 
wW = 320'1b.| per sq. ft. 
Without Surcharge fac: 






------- ©---- 20 Ft, 0" ----------- 





SB age = 





With Surcharge 


Fra. 205.—Details of Basement Wall Supported on Top and Bottom. 
(See p. 641.) 


WALL SUPPORTED AT THE COLUMNS 643 


Wall Slab on the Railroad Side.—The slab is subjected to a surcharge pres- 
sure of 6 ft. The earth pressures are represented by a trapezoid shown in Fig. 
205. At the top, wi = 180 lb. per sq. ft., and at the bottom, w. = 500 Ib. per 
sq. ft. 

The bending moment due to a loading in the form of a trapezoid may be 
found by Formula (9), p. 640. M = 0.75(wi + w:)h? in.-lb. 

It is: 

M = 0.75(180 + 500)10.672 = 58 000 in.-lb. 
58 000 


Pepereeinens Bs TN bs A 2 San 8a in, 
0.875 X 10.75 X 16.000 she 


3-in. round bars, 94 in. on centers, give an area of 0.388 sq. in. 


0.385 Mar : 
The ratio of steel, p = 12 < 10.75 = 0.003, indicates that the stresses in 


concrete are well within the allowable working limits. 


Wall Supported at the Columns.—When the basement wall is 
provided with wide window openings below the first floor level, or 
when bulkheads are provided in the slab, the wall cannot be con- 
sidered as supported at top and bottom, and it may be economical 
to span the wall between the columns. The bending moments 
produced by the earth pressure will depend upon whether the wall 
is continuous over several spans or is only one span long. 

The main reinforcement is horizontal, and is placed near the 
inside face of the wall. When the wall is continuous, negative rein- 
forcement at the column near the outside face should be provided. 

Some vertical temperature steel is commonly used. This may 
consist of 2-in. round bars spaced 18 in. on centers. Sometimes, 
additional bars of larger diameter are used to support the horizontal 
bars during construction. 

Bending Moment in Wall Slab.—If the ground is level, the earth 
pressure for any horizontal strip of wall is constant. The bending 
moments may be computed in the same way as for floor slabs. 
Since the intensity of earth pressure varies with the depth, the 
bending moments will be greatest for the lowest strip and will 
decrease toward the top. The thickness of the wall is usually con- 
stant for the whole height; therefore, with the decrease of bending 
moment, the amount of steel may be decreased. 

Providing for Negative Bending Moment.—The negative bending 
moment in the wall slab may be provided for by bending one-half 
of the horizontal bars and extending them to the adjacent span, as 
in continuous slabs. It is often difficult to handle long bent bars, 


644 BASEMENT WALLS 


especially on account of the interference of column steel, behind 
which the ends of the bars would have to be placed. The construc- 
tion is simplified by keeping the positive reinforcement straight and 
by using, for negative reinforcement, short horizontal bars at the 
column near the outside face of the wall. 

End Spans of Wall.—In end spans, the amount of steel should be 
increased by 20 per cent. 

Bending Moment in Column.—The earth pressure transferred 
from the slab to the column produces bending in the columns. The 
earth pressure resisted by the column may be represented by a 
triangle or a trapezoid, and its magnitude equals the pressures on the 
wall multiplied by the span. In high buildings with heavy basement 
columns, the stresses in the column due to earth pressure are small, 
and do not have to be specially provided for. For heavy earth pres- 
sure and for light columns, the earth pressures may increase the 
stresses considerably. An inexperienced designer should compute 
the stresses in-columns due to earth pressure in all cases, until he 
develops sufficient judgment. 

The tensile stresses due to earth pressure in the center part of the 
column act at its inside face. When the column extends above the 
first floor, negative moments are developed at the junction of the 
column and the floor. The tensile stresses produced by this bending 
moment act at the outside face of the column. They should be 
added to the tensile stresses produced in the wall column by the 
bending moment transferred to the column from the floor construc- 
tion. 

Earth Pressure.—The variation of earth pressure along the ver- 
tical section of the wall is shown in Fig. 646. For a wall without 
surcharge, it varies as a triangle; and for a wall with surcharge, as a 
trapezoid. Along a longitudinal section of the wall, the pressure is 
constant. The wall is then considered as a slab loaded with uni- 
formly distributed load along its span. The magnitude of this 
uniform load increases with the depth. The load and the bending 
moments will be a maximum at the bottom and will decrease to a 
minimum at the top, in the same ratio as the pressure on a vertical 
section. Since the thickness of the wall is the same throughout, the 
amount of steel required to resist the earth pressure will vary accord- 
ing to a triangle or trapezoid. After the area of steel at the top and 
bottom has been computed, the amount of steel at intermediate 
points may be obtained by interpolation, or graphically as in Fig. 


EXAMPLE OF BASEMENT WALL 645 


206. The varying amount of steel is provided by varying the spacing 
of the bars. When the spacing becomes too large, bars of smaller 
diameter should be used. 


Example of Basement Wall Supported by Column. 


Example 2.—Design basement wall, with columns 3 ft. wide and 2 ft. thick, 
spaced 20 ft. on centers. The distance from top of basement slab to top of first 
floor is 10 ft. Ground level is 4 in. below the level of the first floor. There is a 
continuous opening in the first floor slab, along the wall, for a bulkhead. On 
one side is a railroad track producing a pressure equivalent to a surcharge of 6 ft. 
(Compare the solution of this problem with that on p. 641.) 

Solution.—The wall cannot be supported at the top on the floor slab, because 
of the opening for the bulkhead. It will be considered as spanning between the 
columns. 

Span.—The distance, center to center, of columns is 20 ft. Since the column 
is 3 ft. wide, the net span is 17 ft. The design span, taken as the net span, 
equals 1; = 17ft. (See p. 277.) 

Wall without Surcharge——The maximum unit pressure at the bottom is 
w = 305 Ib. per sq. ft., as is evident from Fig. 206. The bending moment for 
continuous wall with 17-ft. span and 305 Ib. load, from formula M = wi? in.-lb. is, 


M = 305 X 17? = 88 100 in.-Ib. per lin. ft. of height. 


For a 12-in. wall, the effective depth is d = 10.5 in. The amount of steel: 


from formula A, = ——, is 


jdf g 
53 88 100 
~ 10.5 X 0.875 X 16 000 





= 0.60 sq. in. per foot. 


As 


-in. round bars, 6 in. on centers, will satisfy the requirement. 

The spacing of the bars for other sections of the wall is shown in Fig. 206. 
It may be noted that near the top of the wall, where the area of steel required by 
bending moment is small, a larger amount of steel is used to provide for tempera- 
ture and shrinkage. 

Walls with Surcharge.—The maximum pressure on the top is 180 lb. per 
sq. ft.; on the bottom, 485 Ib. per sq. ft. (See Fig. 206, p. 646.) The bending 
moments and the required amounts of steel will be computed at the top and 
bottom. These will be plotted and the amounts of steel at intermediate point 
determined. 

Top. Pressure 180 lb per sq ft. 


M = 180 X 17? = 52.000 in -lb. per lin ft. of height. 
A, = 0.354 sq. in. per ft. 


Use $-in. round bars, 12 in. on centers. 


LAs 





20.60 Sq. In. 


at Depth x 

































oe ET. 


cee ee ee ee 





<j , 
Zi S 
= S % 
. S 3 - 
J i 2%. B 
= {i 8 
S = 3 
a 
Fa 
< ee na pinoy 2 
an cas aN es = = 





ee ee 


Area Required 
' at Depth x 
0.95 Sq. in. 


s 


646 
ee 
‘A 





-Out to Out of Building 















” 
gs 
9» 
6 
> 
eS ‘ z 
x Ss 
$ cite oa 
S io? 
~ x : 
pee i eae K ' 
19 if lor 
rs bs i: 
> 7 
a 2 “be 
Ss zi 
< we 
tous a 
I 5 pal " | 
I 1) >| 1 
~ 
| Ss | ! 
\ S { | 
= \ | 
= 1 
= f 
= eS 4 
S 1 
al 
= > 
aS ‘ 
~~ 
= w 
N 


With Surcharge 


a 


485 1b. per|sq. ft: 


Areas 
of Steel 


Section 


ToOOCUML 


PREECE 


SE 











ia 








peed a= =~ 20 Ft, 0" - = 


Fra. 206.—Detail of Basement Wall Supported at the Columns. 





(See p. 645.) 


BASEMENT WALLS FOR AREAWAYS 647 


Bottom. Pressure 485 lb. per sq. ft. 
M = 485 X 17? = 140000 in.-lb. per lin. ft. of height. 
A, = 0.95 sq. in. per ft. 

Use 3-in. round bars, 4 in. on centers 


0.95 


ee OTS. 
PMI GR Se 12 


Since the per cent of steel is smaller than the value corresponding to a compres- 
sion stress in concrete f, = 650 lb. per sq. in., the thickness of wall is satisfactory 
for compression. 

The required area of steel and the arrangement at intermediate points may 
be seen in Fig. 206, p. 646. 


Basement Walls for Areaways.—The design of walls for areaways 
depends upon conditions. 

Walls for small areaways, one panel wide, may be considered as 
supported on the cross walls, and reinforced with horizontal bars 
placed near the inside face. 

If the areaway extends over several spans, it is often possible to 
run struts at the top of the wall to the columns. The wall can then 
be designed as supported on top and bottom. Enough longitudinal 
steel must be used on the top to transmit the pressure to the struts. 
Such construction is shown in Fig. 265, p. 754. 

If struts are not permissible and it is not possible to support the 
top of the wall, it must be designed as a cantilever wall. The base 
must be made strong enough and rigidly connected with the wall. 
The principle of the design is then similar to that explained for retain- 
ing walls. _ The base may be placed either outside the wall or between 
the wall and the building. 


CHAPTER XIII 
ROOF CONSTRUCTION 


Loading.—Roofs should be designed for the dead load, the live 
load, and, in case of inclined roofs, for the wind pressure. 

The Dead Load consists of the weight of the roof construction, of 
the cinder fill or tile placed on the top of the roof, and of the roof 
covering. The unit weights of the various materials used in the 
construction of the roof are given in the table below: 


Dead Loads on Roof 











| Weight, 
Description Lb. per 
Sq. Ft 
Roof Covering: 
Five-ply felt and gravel. 2.....:... 2.50 +: >: 6 
Four-ply felt, and grayel............9->/24 +) se eee ee 54 
Tne «bcc bie cc din wb bus kin ove pom Web SOO ie cc ett ce ann een if 
Metal covering. oc sca wt. ce ee bee = nti oe 5 anne 13 to 2 
Slate 8, to 2-in. thick. . 220.2 s 0 eb eet es 7.25 to 9.5 
Insulating Materials: 
Cinders, per in. ..... 5. si ees se et eee oe oe #f 
Average weight of cinder fill............--++++seneeeeveseees | 70 
Cork or lith insulator... 205 S200 2. 0 0 10 
Tiles... cs «cc wu pins c unumrn :loppilcus Sgehensleets le (eto. ch Sains iain 12 to 25 
Plastering on ceiling on concrete or tile.......----++++.++2255: 5 
Suspended ceiling without ornamentation. .........+++-+++-++5 10 





The Live Load for which roofs are ordinarily designed is intended 
to take care of the weight of snow, and also of the weight of men 
walking upon the roof. Most building codes require a live load of 
AO Ib. per sq. ft. for flat roofs or roofs with a pitch of less than 20 
degrees with the horizontal. For roofs with a larger pitch than 

648 


DRAINAGE 649 


20 degrees, the live load may be reduced to 30 lb. per sq. ft. In 
localities having severe winters, such as the northern part of the 
United States and Canada, the live load may have to be increased 
to 50 lb. per sq. ft. to take care of the additional weight of snow 
and ice. 

Wind Pressure.—Pitched roofs, when the pitch exceeds 20 degrees 
with the horizontal, should be designed for wind pressure, which 
should be assumed to act horizontally, with an intensity of 30 lb. 
per sq. ft. of vertical projection of the roof surface. In determining — 
stresses in the roof members, this horizontal pressure should be 
resolved into pressures normal and tangent to the roof. It is usually 
considered that, with a wind pressure acting on one side of the roof, 
the live load is applied only on the lee side, as it is scarcely possible 
that with wind force of 30 lb. per sq. ft. any snow would remain on 
the side exposed to the wind. 

Drainage.—A flat roof should be provided with a sufficient 
number of outlets to carry off the rain water. These should be 
properly distributed over the roof. To facilitate the running off of | 
water, it has been customary to pitch the tops of the roofs. Lately, 
the necessity of pitching the top is questioned, with much justifica- 
tion, as with properly distributed outlets, most of the water will run 
off without any pitch. Some water may remain on the roof until it 
dries out, but this does no harm to the roofing. The only advantage 
of pitching the roof, therefore, seems to be that in case of cracks in 
the roofing the leakage through the slab would be smaller with 
pitched roof than with flat roof. With reliable roofing, the chances 
of leakage are remote. Therefore, the authors recommend dead 
level roofs when high-grade roofing is used. 

Pitch of Roof——The customary pitch for gravel roofs is 2 in. 
per lineal foot. For smooth tile, this pitch may be reduced to # in. 
per foot. Pitching of the roof is accomplished either by pitching 
the whole roof construction or by building the roof slab horizontally 
and providing a sloped fill on the top of the roof slab. The method 
to be adopted will depend, first, upon architectural requirements 
(i.e., whether there is any objection from the architectural stand- 
point to an inclined ceiling), and second, upon the method of pre- 
vention of condensation. ‘Thus, if cinder fill is used on the top of 
the roof to prevent condensation, this could be utilized to give the 
required pitch. This also applies where light construction is placed 
above the roof to provide air space, 


650 ROOF CONSTRUCTION 


_ The simplest method of pitching a roof is to raise one edge of the 
building and make one continuous pitch across its whole width. 
This method is particularly advantageous where the roof slab is 
pitched, as the building of the formwork is simpler than for more 
elaborate pitching. It may be used only in narrow buildings; 
otherwise, the difference in story height on two opposite sides would 
be large. For wider buildings, a ridge or a valley may be provided 
along the center of the building. With large roof areas, considerable 
thought must be given to the method of pitching. First, the number 
of outlets should be computed. These should then be placed in the 
most suitable positions, and the pitch in the roofs provided accord- 
ingly. Two or-three trials may be necessary to give the simplest » 
solution of the problem. It should be remembered that the down- 
spouts are best placed near columns, thereby restricting the number 
of possible positions for the outlets. 

Figure 207, p. 651, shows the location of the outlets and the 
method of pitching adopted in building. 

Roof Insulation.—Concrete roofs often require insulation to 
prevent condensation and loss of heat through radiation, in cold 
weather. Insulation of roofs is required in all buildings used for 
habitation, office buildings, hotels, hospitals, and schools. In ware- 
houses, and in some cases in factories, solid concrete roofs covered 
with tar and gravel roofing require no additional insulation. 

By condensation is meant the accumulation on the ceiling of 
moisture from the air, caused by a difference between the tempera- 
ture of the concrete roof and that of the air in the building. To 
each inside temperature and degree of humidity there corresponds a 
definite temperature of the roof at which condensation will occur. 
This temperature is called the dew point. The object of insulation 
is to prevent the outside temperature from reducing the tempera- 
ture of the inside surface of the roof to the dew point at the time when 
the conditions for condensation are most favorable, that is, when 
the difference between the inside and outside temperatures is greatest, 
and the humidity, or moisture content of the air, is also greatest. 

The condensation depends upon the following factors: 

(1) Greatest difference between outside and inside temperatures. 
Condensation increases with increase in difference. 

(2) Expected moisture content or humidity at the time of greatest 
difference in temperature. Condensation increases with increase of 
humidity. 


651 


ROOF INSULATION 


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652 ROOF CONSTRUCTION 


(3) Degree of conductivity of heat of the roof construction. For 
instance, for an 8-in. slab, the condensation will be smaller than for 
9-in. topping in concrete joist construction. 

Humidity may be atmospheric or it may be produced by some 
process of manufacture. Atmospheric humidity can usually be 
taken care of without difficulty by proper insulation. Special 
humidity may require special ventilation in addition to insulation. 
In some cases, complete prevention of condensation is impossible. 

Methods of Roof Insulation.—There are several methods of roof 
insulation, and the choice between them will depend upon relative 
economy. | 

The methods are: (1) Providing dead air space between the 
ceiling and the roofing; (2) covering the roof slab with cinder fill; 
(3) covering the roof with insulators like fibrofelt, lith, or cork; 
(4) placing on top of the roof hollow tile either of clay or gypsum, 
(5) combination of tile with cinder fill or with fibrofelt. (The order 
in which these methods are arranged has no significance.) 

In determining the relative economy of various types of insula- 
tors, the dead load of the insulator also should be considered. For 
instance, the dead load of cinder fill amounts to 70 lb. per sq. ft., 
while the dead load of cork insulator or lith amounts to about 10 lb. 
The additional dead load requires stronger roof construction, also 
heavier columns. ‘The additional cost of construction must be added 
to the cost of the heavier insulator. 

Method of Providing Air Space.—The best method of providing 
dead air space between the roof and the ceiling is by a suspended 
ceiling, such as that described on p. 613. Air space of any dimen- 
sions may be thus obtained. 

A much less effective method, used in beam and girder and joist 
constructions, is by providing a plastered ceiling directly under the 
beams and joists, as described on p. 610. The airspace thus pro- 
vided is limited by the height of the beam or the joist. In addition, 
the joists are not insulated. 

A cheaper method of providing air space above the ceiling, used 
to a great extent in Canada, consists of a light wood framework built 
some distance above and supported by the concrete slab. Usually, 
this consists of 2 by 4 in. planks placed and anchored to the concrete 
slab, and vertical wooden studs resting on the planks and spaced 
about 4 ft. on centers. Stringers supported by the studs carry rafters 
upon which rests tongue and grooved planking covered with appro- 


INSULATORS 653 


priate roofing material. Proper pitch can be easily obtained. An 
objection to this method is that the roof is not strictly fireproof. If 
the roofing is fire-resistant this is not a serious objection. 

A small air space may be obtained by placing wood strips of 
proper thickness, spaced from 12 to 18 in. apart, directly on the con- 
crete. On these rest the boards, and on top the roofing. 

Cinder Fill.—Cinder fill placed on the top of the roof slab is the 
most commonly used method of preventing condensation. The 
concrete slab is built level, and the desired pitch in the roof is pro- 
vided by varying the thickness of the cinder fill. To be effective as 
insulator, the fill in the valleys should be at least 6 in. thick. The 
slopes should be so arranged that the thickness of the fill at high 
points is not excessive. It is desirable to limit the average thickness 
of the fill to 12 in., in which case its average weight will be about 
70 Ib. per sq. ft. 

The cinder fill is made of wet hard cinders properly tamped and 
covered with a mortar coat one inch thick. The roofing is placed on 
the top of this coat. To provide for expansion of the cinder fill, a 
one-inch joint, filled with asphalt or other bituminous material, is 
usually provided at the parapet. This is particularly important 
with brick parapets, because, if the joint were-not provided, the 
thrust due to expansion of the fill might injure the parapet, or, on 
the other hand, contraction cracks might allow penetration of water. 

Insulators.—Roofs may be insulated by placing, on the top of 
the concrete slab, insulators such as fibrofelt, lith, or cork board, and 
covering them with roofing. The fibrofelt comes in various thick- 
nesses up to 13 in. Lith and cork board come in sheets. Lith may 
be obtained in thicknesses up to 4 in. Both materials can be sawed 
like wood, to fit the roof. 

A coat of pitch is placed first, and the insulator is placed on the 
top and covered with another coat of pitch, upon which the roofing 
is applied in the regular manner. These materials show considerable 
resistance to loss of heat. 

Tile Insulators.—Lither clay or gypsum hollow tile may be used 
for roof insulation. When erected, the tiles are placed end to end 
so as to form a continuous air space. Gypsum tile offers larger 
resistance to loss of heat than clay tile, but under some conditions 
it deteriorates and loses its insulating properties, particularly when 
it is exposed to heat or when damp air can penetrate to it. 

When tiles are used as insulators by themselves, the roofing is 


654 ROOF CONSTRUCTION 


placed directly on top of the tiles. Sometimes they are used in con- 
nection with cinder fill, fibrofelt, or cork. In such cases, the tiles 
are placed on the concrete and the additional insulator is placed on 
the top. 

Roof Covering.—Concrete roof slab may be made impervious to 
water without any special covering, by the use of properly propor- 
tioned concrete reinforced with sufficient temperature reinforcement 
and troweled to a dense and hard surface. Since this is not a positive 
insurance against leakage, in building construction the roofs are 
usually covered with waterproof roof covering. Concrete roofs are 


_...Flashing 
Form 





LQos 


Fic. 208.—Method of Application of Tar and Gravel Roof. (See p. 655.) 


used in reservoirs and structures of similar character where dampness 
of the under surface and occasional leakage are not objectionable. 

The roof covering materials may be divided into three classes: 
namely, roof coverings relying for waterproofing on coal tar or 
asphalt; roof coverings consisting of slate or tile; and metallic roof 
coverings, such as tin, copper, and corrugated iron. 

Tar and Gravel Roofing.—For substantially flat roofs, the tar and 
eravel roofing is used very extensively, as it gives a lasting roof 
surface at comparatively low cost. For roofs with large pitch, tar 
and gravel roofing cannot be used, as there is danger of the roofing 
running during hot weather. The maximum pitch for which this 


ROOF COVERING 655 


roofing may be used depends upon the climate. In hot climates, it 
should not exceed one inch to the foot, and in cold climates three 
inches to the foot. 

The tar and gravel roofing consists of three materials, namely, 
pitch, felt, and gravel or slag. The pitch provides the waterproofing 
qualities; the felt serves to keep the pitch in place; while the gravel, 
which is imbedded in pitch on the top of the roof covering, serves as 
a protection for the roofing both against injuries by physical action, 
like walking and scraping, and also against the action of the direct 
rays of the sun. 

For satisfactory results, the materials and the workmanship 
must be of first-class quality. The method of application is as fol- 
lows: The pitch is spread ar 
first on the concrete; then | Fu Plies of Felt Ritchie. 
the required number of '(7¥ a ae, 
layers or plies of tarred \ CO aa! & al Af ee LA AO 
felt are placed. Thecearc —>E————————————— 
lapped in such a fashion 
that at each point of the 
roof there are the same 
number of layers. The : 
tarred felt must be covered Four heavy moppings of Pitch” 




















with pitch, so that in no Fig. 209.—Section through Tar and Gravel 
place felt touches felt. The  . Roofing. (See p. 655.) 


felt must be spread without 
wrinkles, as these destroy the waterproofing qualities and contribute 
very largely toward breaking of the felt. 

This method is illustrated in Fig. 208, p. 654. The relative 
thicknesses are evident from a large scale section through the roofing 
shown in Fig. 209, p. 655. 

The following specifications, universally known as “Barrett’s 
Specifications,” are recommended for use. 


First—Coat the concrete uniformly with Specification Pitch. 

Second—Over the entire surface, lay four plies of Specification 
Tarred Felt, lapping each sheet twenty-four and one-half inches over 
preceding one, mopping with Specification Pitch the full twenty- 
four and one-half inches on each sheet, so that in no place shall 
Felt touch Felt. 

Third—Over the entire surface, pour from a dipper a uniform 
coating of Specification Pitch, into which, while hot, imbed not less 
than four hundred pounds of- Gravel or three hundred pounds of 


656 ROOF CONSTRUCTION 


Slag for each one hundred square feet. The Gravel or Slag shall be 
from one-quarter to five-eighths inch in size, dry, and free from dirt. 

General—The Felt shall be laid without wrinkles or buckles. 
Not less than two hundred pounds of Pitch shall be used for con- 
structing each one hundred square feet of completed roof, and the 
Pitch shall not be heated above 400 degrees Fahrenheit. 


Slate Shingle and Clay Tile.-—These materials are used on roofs 
with a pitch of more than 6 in. to the foot. Owing to the steepness 
of the roof, bituminous roofing cannot be used, as it would run. | 
The shingles and tiles are nailed to 1 by 2 in. strips, imbedded in 
concrete under each row of shingles. This covering 1s made water- 
proof by proper lapping of the tile at all joints. | 

Metallic Roofing.—This is usually expensive and is seldom used 
in connection with concrete roofs. It has been found by experience 
that the metal lasts much longer if it is placed on planks instead of 
‘directly on concrete. | 

Parapets.—Buildings with flat roofs are usually provided with 
parapets extending above the roof. The purpose of the parapet is 
to give the building proper architectural proportions, to hide any 
drainage slopes, and finally, in non-fireproof buildings, to retard the 
spread of fire from the adjoining buildings. This last purpose need 
not be considered in reinforced concrete buildings. 

The height of the parapet will depend upon the architectural 
design. It may be uniform for the -whole building, or parts of it 
may be higher than the rest. The minimum height of parapets in 
business buildings is 3 ft. for outside walls and 18 in. for party walls. 

The parapet is usually built of reinforced concrete, brick, or tile 
faced with brick. ‘The selection of the material depends upon the 
type of the building and also upon economy. 

Concrete Parapets—Reinforced concrete parapets may have 
exposed concrete surfaces, or the surfaces may be faced with brick, 
stone, terra cotta, or any other suitable material. When exposed, 
the concrete surface may be plain or ornamental; but the ornamen- 
tation, if any, should be such as to require no elaborate formwork. 
‘Concrete cornices are often used. They may either be pre-cast. or 
cast in place. When the shape of the cornice is simple, it may be 
built as a part of the parapet. For larger cornices with more intri- 
cate design, it is more economical to build the projecting parts sepa- 
rately from the supporting concrete parts. Proper provision, 
usually consisting of proper recesses ‘and steel anchors, should be 


PARAPETS 657 


made to support the separately built parts. When the surface of the 
concrete is exposed, sufficient temperature reinforcement should be 
provided to prevent shrinkage and temperature cracks. 

When the parapet is faced with brick, an angle of proper dimen- 
sions should be used at the bottom to retain the face brick. 

When terra cotta is used, proper anchors should be provided for 
attaching terra cotta blocks. 

From a structural standpoint, the parapet may form a part of 
the supporting structure, i-e., it may form a beam supporting the 
roof load, in which case it should be designed to carry the loads 
coming upon it, according to the methods recommended for other 
concrete beams. In case the roof beam consists of a portion below 
the roof and a portion above the roof, it is best, from a structural 
standpoint, to have both parts built at one pouring. The design of 
the beam is then similar to that of any other beam. If, however, it 
is found economical to build the part above the roof separately, © 
special provision should be made to insure the cooperation of the 
two parts. This is usually accomplished by providing stirrups 
throughout the whole length of the beam, extending from the bottom 
portion to the top portion. With stirrups spaced closely enough, 
it is possible to obtain a beam that has the same strength as if it 
were poured at one time. In case of heavy loads, however, it may 
be advisable to increase the required amount of reinforcement, to 
provide for the possibility of incomplete cooperation of the two 
parts. : 

When a concrete parapet supports a heavy cornice projecting 
some distance outside of the building, it is subjected to bending 
moments produced by the weight of the projecting cornice. This 
bending moment equals the weight of the cornice multiplied by the 
distance between the center of gravity of the cornice and the center 
of the parapet. In some cases, it is possible to resist this bending 
moment by vertical reinforcement in the parapet, placed near its 
inside face. The amount of such reinforcement equals the bending 
moment divided by effective thickness of parapet. The maximum 
stress in this reinforcement acts at the junction of the parapet and 
the roof construction, and the reinforcement must be properly 
anchored in the roof construction. The roof members in which the 
bars are anchored must be strong enough to resist the bending 
moment transferred from the parapet. Sometimes, however, the 
parapet wall is not sufficiently strong and special provision, such as 


658 ROOF CONSTRUCTION 


special brackets or extension of the columns above the roof, must be 
made to carry the cornice. 

When the parapet does not serve any structural purposes, i.e., 
when the beam below is sufficient to carry the roof load, it may be 
built separately. It then requires only temperature reinforcement 
and dowels joining the parapet to the roof. The thickness of the 
parapet should be at least 8 in. For very high parapets, this thick- 
ness may have to be increased to 12 in. 

At the parapet a cant, consisting of cinder concrete, should be 
built to prevent accumulation of water at the wall. A perspective 
drawing of a concrete 
parapet is shown in Fig. 
210, p. 658. 

Better results are 
obtained by using special 
continuous reglets made 
of steel. These are 
tacked lightly to the 
inside of the forms 
for the parapet. Such 

Fic. 210.—Concrete Parapet. (See p. 658.) reglet is shown in Fig. 

210, p. 658. 

Proper provision should be made, on the inside face of the para- 
pet, for attaching the flashing. For this purpose, there must be a 
longitudinal groove along the parapet, which may be easily obtained 
by nailing to the form a strip of wood of proper dimensions. 

Coping.—The coping for a reinforced concrete parapet may be 
built at the same time as the rest of the parapet, or it may be built 
separately. In the last case, special reinforcement consisting of 
three 3-in. bars should be used in coping. 

Brick Parapet.—A brick parapet should be at least 12 in. thick. 
It should be built of hard brick, laid in Portland cement mortar. If 
a cornice of any size is used in connection with a brick parapet, it 
should be anchored to the concrete frame rather than to the brick 
parapet. ‘To prevent any possibility of displacement of the parapet, 
either proper recess should be provided in the concrete or else steel 
dowels should be extended up from the concrete into the brick. 

To provide a slot in the brick for fastening the flashing in the 
connection, a special tile built with a groove of proper dimensions 
may be used to advantage. 





ROOF DESIGN 659 


The parapet wall should be provided with a coping of stone, 
terra cotta, concrete, or cast iron. The inside face of the parapet 
should be coated with tar, so as to prevent moisture and frost from 
penetrating the masonry. 

Walks on Roof.—Where considerable walking is done on the reof, 
special walks should be provided to protect the roofing. A simple 
construction consists of 2 by 3 in. sleepers placed directly on the roof 
across the walk, with 1 by 2 in. slats, spaced 1 in. apart, laid over the 
sleepers. 

‘Where the roof is used for a promenade, it should be surfaced 
with lasting material, such as vitrified tile. 

First the roofing is placed as usual; then cement mortar is placed 
on top, and the tile is imbedded in the mortar. Expansion joints 
should be provided in the mortar, to prevent injury to roofing. 

Flashing.—F lashing is provided to make the junction of the 
parapet and the roof waterproof and to prevent water from entering 
under the roofing at the parapet. There are several varieties of 
flashing in use. | 

Metal Flashing.—This may be of tin, galvanized iron, copper, or 
lead. It is usually nailed at the bottom to a nailing strip imbedded 
in the slab a proper distance from the parapet usually from 6 to 
10 in. The vertical portion of the flashing is placed in a specially 
prepared groove or reglet in the parapet, the groove being filled with 
cement mortar after the flashing is in place. To prevent the mortar 
from coming out, the groove is tapered. The method of procedure 
is as follows: The felt is applied first, extending to the parapet. 
Then the metal is nailed on the top. Felt is then placed over the 
horizontal part of the flashing; and finally a coat of pitch and gravel 
is applied. 

Elastic Flashing.—Figure 211 shows the flashing advocated by 
the Barrett Company. No metal is used. The felt and pitch are 
brought to the edge of the parapet. 


ROOF DESIGN 


In buildings consisting of several stories and roof, all columns 
usually extend to the roof, so that the size of the panels in the roof 
is the same as the size of the panels in the floor. The difference 
between the floor and the roof, from a structural standpoint, is only 
in the magnitude of the loading. It is most economical to use for 


660 ROOF CONSTRUCTION 


the roof a design of the same type as for the floor, so as to be able to 
use the formwork from the floors below. ‘Thus, if the floors are of 
flat slab design, the roof also will be a flat slab. If the floors consist 
of beams and girders, the roof will be of beam and girder design. 

The loading for which the roof 1s designed is usually smaller 
than the floor load; therefore, beams of smaller dimensions may be 
used in the roof. Before deciding, however, upon a change in the 
size of the concrete beams, the cost of the changes in formwork must 
be seriously considered. In many Cases, the saving in concrete 
- materials, obtained by using smaller concrete dimensions in the roof, 
‘s more than offset by additional cost of the changes in formwork. 
A change in the amount of steel, instead of a change in the size of 
concrete members, will often accomplish the desired result more 
economically. If it is desired to change the dimension of beams, 
the depth of the beam, rather than its width, should be reduced. 
A reduction in width of the beam requires not only changes in the 
beam formwork, but also changes in the slab formwork, to provide 
for the greater distance in clear span between beams. The thickness 
of the slab can be reduced without affecting the formwork. It 
should be noticed that this automatically reduces the depth of the 
beam. Note that it is not the total depth of the beam, but the depth 
below the slab, which affects the formwork. In flat slab construction, 
a change in thickness of slab does not affect the formwork. Form- 
work needs to be changed only when the depth of the drop panel is 
reduced, and this change is very inexpensive. 

Sometimes it is desired to get larger spans in the top floor than 
in the lower floors, by omitting one or more rows of columns. In 
such cases, the entire design of the roof is changed. In beam and 
girder construction, the cross beams will be of a much longer span, 
requiring special concrete dimensions and, therefore, special form- 
work. The same spacing of the longitudinal beams, however, should 
be used as in the lower floors, so as to utilize at least a part of the 
formwork. When the lower floors are of flat slab design, the design 
of the roof must be altogether changed, and such arrangement of 
- beams used as will give greatest economy. 

Openings in Roof.—lIn buildings intended for manufacturing 
purposes, it is often desired to get additional light and ventilation 
in the top story by the use of skylights or monitors. 

With flat slab construction, the openings for skylights should be 
arranged so as to require the least amount of framing. If the sky- 


LONG SPAN ROOF CONSTRUCTION 661 


light is placed in the central portion of the panel, it may be possible 
to get along without any framing. When framing is required, it 
consists of two beams at the edge of the skylight, supported by two 
girders running between columns at right angles to the beams. 

To make the connection of the skylight with the roof weather- 
proof, it is advisable to place the skylight some distance above the 
highest point of the roof at the skylight. This is accomplished by 
providing, around the opening above the roof slab, a concrete curb 
6 in. thick and of proper height. The curb is usually built after.the 
slab is completed, and its height depends upon the height. of the 
cinder fill at the skylight. To join the roof and the curb, dowels are 
provided, extending from the roof into the curb, and a water-tight 
joint must be made by clearing the old surface and spreading upon 
it neat cement parts. Flashing is extended from the roof to the top 
of the curb, where it is nailed to wooden nailing blocks provided 
for this purpose. (See Figs. 211 and 212.) The nailing blocks 
should be imbedded in concrete before it is set. 


Metal apron set in place after 
Slashing has been nailed to top 
of curb 

Flashing run up over top of > 


curb and nailed to continuous _ Be _--Slate Surface Roofing 
creosoted wood nailing strip ~~ 





~~ Three layers of roofing felt 
with alternate layers of Elastigum, 
ee run up the curb and out on 
we roofing felt at least 6" 








All plies of the Roofing felt 
are cut off at the angle of ~~~... 
curb and roof deck os 






“~--> Roofing... 


nr, 
y FE aR aS eg OSs gts aN Tn COs eye intel Daae piltiia May 
ES: ae RRNA Rane ante = ocr iS ie 3.9. a wdieita 


Fig. 211.—Flashing at Skylight, (See p. 661.) 
LONG SPAN ROOF CONSTRUCTION 


In long span roof construction, concrete is seldom as cheap in 
first cost as light steel truss construction, but it has the advantage 
of being fireproof and of requiring practically no expense for main- 
tenance. ‘The appearance is also in favor of concrete construction. 
The maintenance chrages for steel trusses are fairly high, as they 
need to be regularly repainted. Where steel trusses need to be fire- 
proof, concrete construction is more economical than steel con- 
struction. 

Several types of long span roof construction are in general use. 
These are simple girders, concrete trusses, rigid frames, and long 


662 ROOF CONSTRUCTION 


span arches. Of interest also is the long span roof construction 
developed by Richard E. Schmidt. (See p. 671.) 

Simple Girder.—Simple girders have been used to quite an 
extent for long spans in garages and manufacturing establishments. 
They are usually supported by light concrete columns or by brick 
piers. The light concrete columns are seldom capable of offering 
appreciable restraint to the girders. 

Thero of construction may consist of girders 18 to 20 ft. apart 
supporting transverse beams spaced about 8 ft. and carrying a 
2-in. concrete slab. Light weight concrete construction, as de- 
scribed on p. 602, may be substituted for the transverse beams and 
slab. To save formwork for intermediate beam the spacing of 
girders may be made smaller and slab spanned between them as 








If open joints occur between 
angle sills they should be sealed 
with Elastigum and capped 
with metal 


Angle bolted in place after flashing 
has been nailed at top of curb 


/ Slate Surfaced Roofing 


Three layers of roofing felt with 
alternate layers of Elastigum, run 
7 up the curb and out on roofing 
felt at least 6” 


Flashing nailed to creosoted-=__ 
wood nailing strip ae 


All plies of the roofing felt 
are cut off at the angle of-------- 
curb and roofing deck 


Fic. 212.—Flashing at Monitor. (See p. 661.) 


shown in Fig. 213, p. 663. The girders are usually of T-shape 
design. For long span girders to resist compression stresses, the 
required thickness of the flange may be larger than the 3-in. slab 
used in the roof, in which case the slab is thickened on both sides 
of the girder sufficiently to furnish the required compression area. 
To reduce the cost of the girder and particularly its dead load, 
it may be economical to make the thickness of the stem in the 
middle portion of the girder, where the external shear is small, only — 
large enough to accommodate the longitudinal reinforcement, and 
to increase it gradually towards the supports with the increase of 
the external shear. Before adapting such a design, it is necessary 
to investigate the cost of the concrete saved and the extra cost of 
the formwork. Since the dead load of the girder is reduced, there 
will be an additional saving in steel, due to the reduced bending 
moment, and also a saving in the size of supporting members. 


663 


LONG SPAN ROOF CONSTRUCTION 


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664 ROOF CONSTRUCTION 


For the sake of appearance, it is advisable to provide in the 
center of the girder a camber of a depth at least equal to the total 
expected deflection of the girder. The objectionable appearance of 
-a sag is thus avoided. 

From the standpoint of design, long span girders do not offer any 
special difficulty. The bending moment is found from the formula 
for simple beams, the distance from center to center of the support 
being taken for the span. In computing the amount of steel and the 
compression in concrete, it may be advisable to use the formulas in 
which the compression in the stem of the T-beam is considered. T he 
simplified method given on p. 224 will be found of advantage in this 
connection. 

Particular attention is required in spacing the longitudinal rein- 
forcement. As the bars are more numerous than usual, it may be 
necessary to place them in several layers. The use of more than three 
layers is not advocated. Special provision should be made for sepa- 
rating the layers of bars. This may be accomplished by placing 
across the main reinforcement short pieces of one-inch square bars 
3 to 4 ft.on centers. A proper proportion of bars should be bent up 
and utilized as diagonal tension reinforcement. Before bending the 
bars, the required amount of steel at intermediate points should be 
determined. ‘This requirement is particularly important in girders 
in which the restraint at the support is slight. (See Fig. 98, p. 292.) 
The points where reinforcement may be bent up will then be deter- 
mined by bending moment in simple spans, and they will be much 
nearer the support than is the case in continuous beams. The use 
of standards applicable to continuous beams may be disastrous. 
At least one-third of the bars should be carried straight to the end 
of the beam. Under no circumstances should the bars be stopped 
short at any intermediate point, as there is no means, in such cases, 
to develop the stresses in the short bar, without its slipping. 

When the ends of the girder are not free to move, short bars 
should be used on the top of the girder at the support, to insure against — 
cracks, even if the restraint is not counted upon. 

When a concrete girder rests on brickwork, the bearing stresses 
on the support should be investigated. If the weight of the girder pro- 
duces excessive bearing stresses, the bearing area should be increased. 

An example of long span girder construction is given in Fig. 213, 
p. 663, showing the roof girder used in the Winston-Salem Auditorium. 
The location of the girders is shown in Fig. 275, p. 781. 












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Fia. 215.—Details of Foundry Building for Garfield Smith & Co. (See p. 665.) 


(To face page 668.) 


LONG SPAN ROOF CONSTRUCTION ~ 665 


Long Span Roof Girders as Rigid Frame.—The cost of long span 
girders may be materially reduced, by building the girder mono- 
lithic with the supporting columns and designing the construction 
as a rigid frame. The use of rigid frames in America has not been 
as wide as the economy of this type of construction warrants. This 
is partly due to the fact that the design is somewhat complicated 
and that the building codes and standard specifications have no 
provision for such construction. The arbitrary bending moment 
requirements that are often imposed make proper use of rigid frames 
impossible. 

Design of Rigid Frames.—Rigid frames should be designed 
according to formulas given in the section on rigid frames in Volume 
III. These are just as reliable as the ordinary beam formulas, as has 
been proved not only by a number of successful constructions but 
also by numerous tests. 

Rigid Frame with Horizontal Girders.—Figure 214 shows a good 
example of a rigid frame design with horizontal girder. It has been 
used by Thompson & Binger in the construction of a boiler house for 
Hires Condensed Milk Co. in Morristown, N. Y. The span of the 
rigid frame in this design is 60 ft. and the height of the column 
27 ft. 10in. The dimensions are shown in the figure. It should be 
noticed that the flange of the T-shape was obtained by thickening 
the slab on both sides of the girders. Of interest also may be the 
method of splicing of the bars, which was necessary because it was 
not possible to get bars more than 60 ft. long. The frame was 
considered as partly restrained at the bottom. This is an inter- 
mediate case between a frame hinged at the bottom and one rigidly 
fixed there. The assumption was justified by the fact that the 
columns were tied by reinforcement to substantial footings. 

Rigid Frame with Ridged Roof.—An interesting example of 
ridged roof construction may be found in Fig. 215, opp. p. 665, and 
also in the photographs, Fig. 216, p. 666, representing the foundry 
built for Garfield Smith & Company, by Villadsen Bros. The span 
of the frame is 70 ft., the height is 45 ft. The details of the design 
may be seen from the figure. The frames were designed as hinged 
at the bottom. Of interest also is Fig. 216, showing the method of 
erection of the reinforcement and of the formwork. The rein- 
forcement for the columns and the beam was designed so that it 
could be separated into three parts. Each part was assembled on 
the ground and placed in the formwork as a unit. After all three 


666 ROOF CONSTRUCTION 














Fic. 216.—Views of Foundry Building for Garfield Smith & Co. 


LONG SPAN ROOF CONSTRUCTION 667 


parts were in place, additional bars were placed, joining the three 
sections. The formwork was also divided into three sections. 
These were assembled on the ground and lifted into position by a 
derrick. Fig. 216 shows the central part of the form, with the rein- 


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Fig. 217.—Hall Frame at Lausanne. (See p. 667.) 


forcement placed within, being hoisted into position. This ingen- 
ious method undoubtedly contributed largely to the economy of the 
construction. . 

Rigid Frame with Arched Roof.—A good example of this type of 
construction is seen in Fig. 217, representing a frame in a reinforced 


668 ROOF CONSTRUCTION 


hall at Lausanne, Switzerland. This hall is rectangular in shape, 
183 ft. 9 in. (56 meters) long, and 114 ft. 10 in. (35 meters) wide, 
‘out to out. The construction consists of 8 frames spaced 26 ft. 
3 in. (8 meters) on centers. Fach frame consists of an arch sup- 
ported on columns hinged at the bottom. The intrados is a three- 
centered curve, while the extrados is a segment of a circle, the radius 
of which is 82 ft. (254 meters). This makes the depth of the arch a 
minimum at the crown, where ++ ig 3 ft. 7 in. (1.10 meters) and a 
maximum at the springing line, where it is 8 ft. 3 in. (2.5 meters). 
The cross section of the frame is given in Fig. 217. It should be 
noted that the section is hollowed out, not so much to save material 
as to save dead load, thus reducing the bending columns and stresses. 
In America, owing to higher ratio of cost of labor to cost of material, 
this scheme may not be found economical. | 

The hinged effect at the bottom of the columns was obtained by 
building the foundation and the frame separately and employing a 
fan-shaped arrangement of dowels between the foundation and the 
frame. These dowels were effective in preventing any displacement 
of the frame, but were not capable of resisting any bending moments. 
Therefore, the frame can undergo angular movements due to changes 
of temperature and movements of foundation without producing any 
stresses. 

The following assumptions were made: snow load, 16.5 lb. per 
sq. ft. (80 kg/m?); wind load for calculation of the arch, 30.6 lb. 
per sq. ft. (150 kg/m?). A temperature variation of 15 degrees 
Centigrade was allowed. The shrinkage stresses were taken care of 
by assuming their effect to be equal to a fall of temperature of 20 
degrees Centigrade. 


ConcRETE ARCHES FOR Lone Span ROOFS 


Reinforced concrete arches with tie-rods form an economical 
_ type of long span roof construction. Their use has not been as 
extensive as the advantages of the type would warrant, mainly 
because the design is somewhat complicated. The cost of form- 
work, which at first sight seems to be large on account of the 
curvature of the arch, need not be unduly expensive, because the 
‘ntrados of the arch can be made up of a number of straight seg- 
ments. 
1 La Technique Modern, Vol. 13, No. 1, Jan., 1921, pp. 12 to 14, 


LONG SPAN ROOF CONSTRUCTION 669 


Method of Design.—The roof construction may consist of arch 
ribs placed across the building and spaced from 15 to 25 ft. on centers. 
The arch ribs support beams or joists running longitudinally, and 
these in turn support concrete slab. A monitor may be easily built 
on top of the arch. 

The arches may be constructed with the supporting columns, in 
which case the construction becomes a rigid frame with curved top; 
or they may be independent of the supports, in which case the 
horizontal thrust produced by the arch must be resisted by tie-rods 
and the arches may be considered as hinged at the supports. Formu- 
las for two-hinged arches, given in Vol. III, should be used. 

The concrete dimensions of the arch and the required reinforce- 
ment should be computed for a condition of load producing maxi- 
mum bending moments. Such a condition occurs when the live 
load is placed on one-half of the arch. 

The tie-rods, however, should be computed for the maximum 
thrust, which is produced by the live load extending over the whole 
span of the arch. In computing stresses, the effect of the length- 
ening of the tie-rods, due to the horizontal thrust, should be taken 
into account. The assumption in all arch designs is based on the 
span remaining constant. This naturally will not be true in the 
case under consideration, as the lengthening of the tie-rods will 
permit a slight displacement of the supports, correspondingly increas- 
‘ing the span length. 

Wind pressure may also have to be considered in designing the 
arches, especially when the building is exposed and the rise is com- 
paratively large. The arch with tie-rods should be securely anchored 
to one support, but should be allowed to expand and move in the 
direction of the span length on the other support. If this motion is 
not provided for, the tie-rods will not become effective until the 
resistance of the support to sliding is overcome, because no stress 
can be taken by the tie-rods without corresponding increase in 
length. With such construction, no provision is required for temper- 
ature stresses. 

Example of Arch Roof.—Figure 218, p. 670, shows the design of 
-an arch 103 ft. 6 in. over all, with a rise of 17 ft. The spacing of 
arches in the building is 17 ft. center to center. The cross section 
of the arch, constant throughout, is 114 by 24 in. The reinforce- 
ment consists of three 7-in. round bars placed at top and three at 
bottom. These bars were properly lapped and also provided with 


670 ROOF CONSTRUCTION 


Three-eighth-inch round stirrups were spaced 
e intrados of the arch 
To facili- 


hooks at the ends. 
uniformly the whole length of the arch. Th 
was nominally a segment of a circle with a 90-ft. radius. 










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tate the formwork, however, the intrados of the arch was made up 
of straight segments. This substitution is practically unnoticeable. 

A single 2}-in. tie-rod with turnbuckles resists the horizontal 
thrust. At each end of the tie-rod is a nut and bearing plate, as, is 


LONG SPAN ROOF CONSTRUCTION 671 


evident from details, Fig. 218, p. 670. To prevent sagging, the 
tie-bars are supported by five 3-in. hangers. 

The roofing is carried by wood rafters 2 by 10 in., spaced 2 ft. on 
centers, seated in the ribs. The longitudinal skylights are sup- 
ported by concrete beams running longitudinally with the building 
and supported by the arches. 

The design was made by L. J. Mensch of Chicago.? 


Lone Span Roor Construction 


Long span roof construction of reinforced concrete has recently 
been developed and successfully used by Richard E. Schmidt, 
Garden & Martin, Architects and Engineers, of Chicago. A typical 
example of this construction is the armory designed to house the 
Illinois National Guard, Danville, Ill. 

The main features of this design are protected by patent. 

Figure 219, p. 672, and Fig. 220, p. 673, show details of the design 
and an interior view of this armory. The roof is supported by 
monolithically poured concrete trusses with a span of 85 ft. A 
monitor, with continuous steel sash both sides, is formed by canti- 
lever beams extending from the peak of the roof. Roof of monitor 
consists of a 3-in. solid slab suspended from girders at end of these 
cantilever beams, and formed on the curve of a catenary to avoid 
dead load bending stresses. Tie-rods are used as bottom chord for 
the trusses and between ends of cantilever beams, to avoid bending 
stresses in main truss members. 

Roof over gun shed, shown to the left of main truss, is of similar 
construction. Balconies for spectators are built between the main 
roof and the truss over gun shed, also at one end of building. Design 
of this balcony is shown by Fig. 219. 


CONCRETE ROOF TRUSSES 


Figure 221 shows a design of a concrete roof truss used in the 
Graumann Theater in Los Angeles. Other details of this structure 
are shown in Fig. 270, p. 764, and Fig. 273, p. 779. The following 
description is taken from Engineering News Record, July 5, 1923. 

The roof trusses are 126 ft. 6 in. in span, center to center of 
columns, 15 ft. 8 in. high; each truss has 61 tons of reinforcing steel 
and 231 cu. yd. of concrete, and carries a load of 750 tons. In the 

? For full description, see Engineering News Record, Jan. 27, 1921, p. 167. 


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LONG SPAN ROOF CONSTRUCTION 673 


building up of these trusses, the soffit of the truss, with its reversed 
decorative forms, was put in place first; the bottom chord steel was 
next laid out and fastened securely in place. Spacers, made of 
1{ by 1{-in. bars bolted together, placed approximately 10 ft. centers, 
were used to hold this bottom chord steel in place. Cement blocks 
were used to carry the load. Care was taken in placing these blocks 
to see that they were staggered so as to allow for the placing of the 
concrete. The bottom chord consists of 14-in. square deformed 
bars, 65 ft. long; a 6-ft. lap was used at the splices with three #-in, 
U-clamps. Particular care was taken to see that these U-clamps 





Fra. 220.—Interior View of Armory at Danville, Ill. (See p. 671.) 


were securely in place and tight. The locations of these splices 
were all laid out so as to stagger with the splice in the bars adjacent 
to them. At the ends of the truss, the bottom chord bars were all 
hooked around the vertical column bars. Bars in the diagonal 
tension members were laid out so as to work in between each layer 
of the bottom chord. These bars were hooked on both ends with a 
cross bar running through the hooks, the top chord being rectangular 
in shape and reinforced as for a rectangular column. 

For discussion of the use and design of reinforced concrete trusses, 
see p. 769. | 


ROOF CONSTRUCTION 


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SAWTOOTH ROOFS 675 


SAWTOOTH ROOFS 


Where Used.—Sawtooth roofs are used in manufacturing plants 
where an abundance of light is required, but where it is necessary 
to avoid the glare due to direct sun rays and the consequent shadows. 
This is best accomplished by providing windows, facing north, in 
every or every other panel of the roof. The resulting construction 
is called sawtooth roof construction. 

Sawtooth roofs may be built with the windows vertical or inclined. 
The construction of the roof with vertical windows is the simpler, 
and also makes it easier to keep the windows clean. However, the 
vertical window does not provide as much light as an inclined window 
of the same size, and for this reason inclined lights are in more favor 
with architects. The maximum inclination with the horizontal is 
governed by the requirement that at no time of the year direct rays 
should be permitted to enter the room. This angle depends upon the 
latitude and may be expressed by the following formula: 


Let « = inclination of the window with horizontal ; 
a = latitude of the location. 

Then 
a = 113°30-—"a. 


In the vicinity of New York, a = 41° ; consequently, the angle 
of inclination is a = 72° 30’. 

The angle between the window sash and the inclined slab will 
depend upon the desired size of the window and the depth to window 
sash. If possible, the angle should be made 90 degrees as this 
facilitates formwork. 

Types of Sawtooth Roofs.—The sawtooth roof may consist of a 
frame, as shown in Fig. 222; or it may be provided with a horizontal 
tension member which resists the horizontal thrust produced by the 
inclined roof members. All intermediate beams should be parallel 
to the frame. Beams longitudinal with the skylight throw shadows | 
on the underside of the slab, thus reducing the effectiveness of the 
light. 

Details.—Special attention should be paid to details of the 
gutter, to make the windows weather-proof. The bottom of the 
window sash should be at least 18 in. above the gutter, on account of 
the possibility of snowdrifts. The gutter should be well drained. 
Down-spoutg should be placed at frequent intervals. A gangway 

lan 


676 ROOF CONSTRUCTION 







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Fig. 222.—Details of Sawtooth 


PRE-CAST CONCRETE ROOF TRUSSES 677 


at least 2 ft. wide should be provided to facilitate the cleaning of the 
gutters. To prevent ice from closing the down-spouts, the under 
side of the gutters should be heated. 

The details of roofing are shown in Fig. 223. 


State surfaced roofing nailed along top of curb 
“into creosoted wood nailing strip 







Pe Three layers of roofing felt with 

/ alternate layers of Elastigum, run 

: / up the curb and out on felt in valley 
at least 6” 


Creosoted wood 
nailing strip~.. 







j-----Steep Roofing 
All plies of valley roofing___ j 
are cut off at the angle ~~ 
Roofing in valley 

nd carried up the slope 


Fig. 223.—Details of Roofing for Sawtooth Roof. (See p. 677.) 


PRE-CAST CONCRETE ROOF TRUSSES 


Pre-cast roof trusses were used for a shed of Pier 6, at Cristobal, 
Canal Zone. The shed is 945 ft. long and 159 ft. wide, and has a 
clearance under trusses of 25 ft. The arrangement of trusses is 
shown in Fig. 224, p. 678. 

The construction is three panels wide. The main trusses, placed 
across the building, are 39 ft. long and 7 ft. deep, and are spaced 45 ft. 
on centers. They are supported by columns. The longitudinal 
trusses, 45 ft. long, 4 ft. deep, and spaced approximately 10 ft. on 
centers, are carried by the main trusses. A 33-in. slab is placed on 
the longitudinal trusses. The depth of slab is small for the 10-ft. 
span and is possible only because there was no snow load to be 
provided for. The design live load was only 10 Ib. per sq. ft. 

The design of the trusses is evident from Fig. 225, p. 679. The 
design of the pre-cast columns is also shown. 

It is evident that the column on the top is provided with recesses 
to receive the cross trusses, as well as those longitudinal trusses 
which frame into the columns. 

All the columns and trusses were pre-cast, and completed before 
being placed in position. Locomotive cranes were used in erection. 
Interesting details of the method of pouring the pre-cast members 
and erecting them are given in Engineering News Record, June 24, 
1920, p. 1232. 


678 ROOF CONSTRUCTION 


AUXILIARY STRUCTURES ABOVE ROOF 


Pent Houses.—The enclosures surrounding the elevator machin- 


extend above the main roof. 


ery and the stair entran 


ce to the roof 


on the walls enclosin 


¢ the pent house. 


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In many cases, a large part of 


the pent-house roof is 
staircase or elevator shaft. 


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Fic. 224.—Partial Plan and Section of Pier 6 at Cristobal, Canal Zone. 
(See p. 677.) 


skylight 1s required (as in staircases well provided with windows) 
the roof consists of a solid concrete slab. 

The main roof construction must be made strong enough to carry 
the pent-house wall and roof. When the walls coincide with the 
outline of the opening, as is usually the case, the beams around the 
opening may be utilized to carry the pent house. Sometimes the 
location of the pent-house wall does not coincide with the location 
of the beams. Such designs should be avoided, because they require 


679 


PRE-CAST CONCRETE ROOF TRUSSES 


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AUXILIARY STRUCTURES ABOVE ROOFS 681 


special beam work for the roof, resulting in a considerable increase 
in the cost of formwork. Sometimes it is possible to reduce the cost 
by placing the beams supporting the pent house on the top of the 
roof slab, and utilizing them as a part of the pent-house wall. These 
beams can be built after the roof slab is completed. 

In some cases, it is desirable to construct concrete framework for 
the pent house and use the walls only as curtain walls. The pent- 
house roof is then supported by concrete wall beams. 

Water Tower.—When the water tank is placed above the roof, 
it may be supported on a concrete water tower. The first cost of a 
concrete tower is larger than that of a steel tower, but it is more 
sightly and requires no maintenance charge. 











Fie. 227.—General View of F actory Designed for Bunte Bros. (See p. 681.) 


In Fig. 226, p. 680, is shown a good design of a water tower used 
in the Clark Biscuit Co., building by Wm. Higginson, Architect. 

A good design of a water tower is shown in Fig. 227, p. 681, in the 
candy factory designed for Bunte Bros., by Richard Schmidt, 
Garden & Martin, Architects. Instead of detracting from the appear- 
ance, as would have been the case with an open water-tower, the 
tower in this design has been changed into an architectural feature. 

The details of the concrete design are shown in Fig. 228. 


CHAPTER XIV 
STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


Stairs serve two purposes: first, they provide a means of exit in 
case of fire; second, they are used for travel from floor to floor. 

In office buildings, tall hotels, warehouses, and factories, equipped 
with passenger elevators, the everyday use of the stairs is com- 
paratively small. In such structures, stairways are required mainly 
for exit in case of fire. This purpose governs their number, location, 
and design. Occasionally, however, there may be considerable local 
movement between two adjoining floors, and special stairways, 
conveniently located, may then be provided. These special stair- 
ways need not be enclosed and do not form a part of the fire exits. 

In private homes, schoolhouses, court houses, theaters, factories, 
and similar buildings without elevator service, the stairs serve 
mainly for everyday use- Their location is governed by convenience 
of movement from floor to floor, and their number may exceed that 
required for fire exits. “The stairs required as fire exits should be 
enclosed. Others do not need fireproof enclosures. ; 

Stairways are relatively expensive to construct. They also take 
up a large amount of space; hence, the tendency to use as far as 
possible. On the other hand, the proper number of stairs in proper 
locations may be of great value in the operation of the plant, or in 
the use of the building. 

No general rule is possible for cases where the stairs are designed 
for constant use. The conditions in each building should be thor- 
oughly studied before the location and number of stairs is decided 
upon. 

Rules for design of stairs as fire exits are more general. They 
will be treated below. 

Fire Exits.—There are two recognized means of exit from a build- 
ing in case of fire: first, horizontal exits through or around a fire 
wall; second, stair exits. Stair exits may be used alone or in com- 
bination with horizontal exits. 

682 


NUMBER OF EXITS REQUIRED 683 


Horizontal Exits.—These are exits through a fire wall, a fire exit 
partition, or a fireproof wall separating two buildings. The exit 
may lead to another building or to another part of the same building. 
It should be provided with a self-closing fire door (1.e., a door nor- 
mally kept closed by some mechanical device) not less than 30 in. 
wide, the actual width depending upon the number of persons for 
whom the exit is intended. Horizontal exit also may be effected 
around a fire wall or a wall separating two buildings, by means of a 
balcony or an exterior bridge connecting two buildings or two parts 
of the same buildings. The balcony or the bridge should be at least 
44 in. wide and constructed of incombustible material. It should be 
enclosed or provided with railings at least 4 ft. high. All doorways 
opening on the balcony or the bridge must be provided with self- 
closing fire doors. All window openings within 10 ft. of the balcony 
- must be protected by fire windows. The purpose of these pre- 
cautions is not only to protect adjoining buildings from fire, but also 
to prevent the entrance of smoke which would make the exit useless. 
There should be no obstruction on the balcony or the bridge. If the 
floors of the connected buildings are on different levels, gradients 
of not more than 1 ft. in 6 ft. should be provided. 

While the horizontal exit is considered more effective than the 
stair exit, its usefulness depends upon conditions over which the 
designer of the building usually has no control. For instance, the 
adjoining buildings may belong to another owner and may not be 
available for a horizontal exit. This prevents their being used as 
often as the stair exits. 

Stair Exits.—Stair exits may consist of enclosed interior stair- 
ways, smoke-proof towers, or outside stairways. 

Number of Exits Required.—The requirements as to the number 
of fire exits in a building differ to some extent in the building codes 
of the various cities. In some codes, such as that of Chicago, the 
number is based upon the area of the building; while in other codes 
it depends upon the number of persons occupying the various floors 
of the building. The latter method is the more logical of the two, 
because if the capacity of each exit is known or assumed, their number 
will be governed by the number of persons who are expected to use 
them. 

The building code, if applicable to the building, should be studied 
first. Where no code applies, the following suggestions may be 
adopted. 


684 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


In all business buildings, at least two fire exits should be provided 
from every floor above the first, irrespective of the number of occu- 
pants of the floor. One of these should be a stair exit, while the other 
may be a horizontal exit. The exits should be arranged so that no 
part of any floor area shall be more than 100 ft. distant from a hori- 
zontal exit or an entrance to a staircase. Additional exits should be 
used if the number of occupants of any one floor served by the exits 
‘exceeds the capacity of two exits. 

In all fireproof buildings that are more than three stories high, or 
occupied by more than fifty persons above the first floor, at least one 
of the stairways should be continuous and in fireproof enclosure. 
This should lead either to the street, alley, or open court, or to a 
fireproof corridor of proper width leading to the open air. 

In all buildings over 90 ft. in height, one of the stairways should 
be a smoke-proof stair tower. / 

Capacity of Stairways.—lt is a well-established principle that the 
capacity of the stairways as fire exits should be such that they may 
accommodate at one time all the persons engaged in the building. 
If horizontal exits are used, the number of persons to be taken care 
of by the stairways may be reduced proportionally. In determining 
the number and the width of stairways, the largest floor area served 
by them should be considered. The persons occupying such floor 
area should find accommodation in the staircase between their floor 
and the floor next below. The number of stairways, where the 
persons are distributed uniformly over each floor area, is determined 
by dividing the total number of persons per floor by the capacity of 
the stairways. Concentrated groups of persons or separated rooms 
may necessitate extra accommodation. 

In determining the capacity of the stairways, it is necessary to 
make assumptions based on judgment. The rules adopted by 
various building codes naturally vary and only designate minimum 
requirements. 

The New York Code allows per person on the stairs, a 22-in. 
width of stairs and one and one-half treads; and on the landing and 
+n the hall within the staircase, 33 sq. ft. of area. If, in addition, a 
horizontal exit is used, stairways need to be provided for only one- 
third of the persons occupying the floor under consideration. In an 
automatically sprinkled building, stairways for only one-half of the 
persons per floor need to be provided. When an automatically 
sprinkled floor is also provided with a horizontal exit, stairways 


ENCLOSED STAIRCASES 685 


need to be provided for one-quarter of the number of persons occupy- 
ing the floor. In no case, however, should the number of fire exits 
per floor be smaller than two, one of which must be a stair exit. 

The National Board of Fire Underwriters recommends a some- 
what simpler rule. The capacity of the stairway per floor may be 
considered as equal to fourteen persons for each 22-in. width of the 
stairway, plus one person for each 3 sq. ft. of the floor area of the 
hallway within the staircase of the floor under consideration and of 
the area of the stair landing between the floor under consideration 
and the next floor below. If the building is provided throughout 
with automatic sprinklers, twenty-one persons may be allowed, in 
the above rule for each 22-in. width of stairways. 

Most of the modern codes agree as to the requirement that a 
width of 22 in. of stairs should be allowed for each person. The 
total width of the stairs must, therefore, be a multiple of 22 in., with 
a& minimum of 44 in. 

Enclosed Staircases.—In fireproof buildings, the staircases used 
as fire exits should be enclosed by fireproof partitions. The doorways 
leading to the staircase should be provided with self-closing fire 
doors. The unobstructed width of the stairs, landings, and hallways, 
within the staircase, should be at least 44 in. If a larger capacity 
is required, the stairs may be increased in width by increments of 
22 in. This means that a minimum width of 22 in. is required per 
person. In a stairway 44 in. wide, two rows of persons can be 
accommodated. Tor three rows, the width must be 66 in., and for 
four rows 88 in. The open side of the stairs must be provided with 
substantial balustrades. 

Hand rails should be provided on both sides of the stairs, and are 
allowed to project into the minimum width, if such projection does 
not exceed 33 in. _ Stairs 88 in. wide should have an additional hand 
rail in the middle. The staircase should be free from any other 
possible obstruction. Winding stairs are absolutely prohibited in 
staircases serving as fire exits. At least one stairway should extend 
to the roof. The New York Code requires that if there are more 
than two staircases, at least two of them shall continue to the roof. 
An appropriate pent house built of brick or concrete should be pro- 
vided above the roof. 

The stairway should lead, in line of direct travel, to a street or 
open alley or to a hallway leading to a street. The hallway must be 
enclosed by fireproof partitions. The width of the hallway should 


686 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


be sufficient to accommodate all who may use the stairways leading 
to it. 

It is important that balconies, used as horizontal exits, be kept 
free. In winter, all snow must be removed from them. 

Scuttles.—If no other provision is made for reaching the roof, 
scuttles should be provided, with a substantial iron ladder fixed in 
place. Scuttle opening should be at least 2 ft. by 3 ft. 

Smokeproof Stair Towers.—In tall buildings, one of the enclosed 
staircases is required to be a stair tower. The height of the building 
requiring this differs in different codes. 

The National Board of Fire Underwriters requires a smoke- 
proof tower or a horizontal exit for buildings over 90 ft. in height 

The New York Code requires smoke-proof towers for. buildings 
exceeding 85 ft. in height. In buildings provided with automatic 
sprinklers, this limit 1s raised to 125 ft. 

A smoke-proof tower is similar in design to an enclosed staircase, 
except that the doors leading from it do not open directly into the 
building but on to an open baleony. The door leading from the 
building also opens on the same balcony. Both doors are self- 
closing fire doors. Such an arrangement is seen in Fig. 229, p. 687. 
The advantage of this arrangement is obvious. The staircase, 
having no connection with the building, cannot be filled with smoke. 


LAYOUT OF STAIRS 


Arrangement of Staircases.—From a structural standpoint, it is 
advantageous to place the staircases in corner panels or wall panels, 
as such location offers the least interruption to the floor construction. 
also, the outside walls are then part of the enclosing walls of the 
staircase. 

When elevators are used, the staircase may be advantageously 
placed next to the elevator wall. Very often, it is possible to accom- 
modate the elevators and the staircase in the same floor panel, thus 
interfering less with the typical design of the floor construction. 
Some of the enclosing walls also serve for enclosing the elevator and 
the staircase. 

In many cases, the staircase and the elevator wall are placed 
outside of the area of the floor, as in Fig. 230. Although this arrange- 3 
ment requires more walls, it simplifies the floor construction. 


SMOKEPROOF STAIR TOWERS 687 











S 
N 
N 


u 
S90“ 








Interior of 
Building 





SSS s4$“s$3$“__o 7° 















ME QA ©a9» 


—_— 





Titi SSS SANS 











Outside Balcony 
Solid Floor 









Plan Elevation 
Smokeproof Tower with Outside Balcony Entrance 















j 
eee 

ae? : 

—7 Vestibule 
Y Opening to - aya ee 

my Ceiling 

Y | 
y Railing- ----}"] | 
g Floor Line 











Elevation 


Railing- “Vestibule Opening 
Plan 


Smokeproof Tower with Vestibule Entrance 


Fig. 229.—Smokeproof Stair Towers. (See p, 686.) 


688 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


In factories, the stairs are placed in the service portion next to 
the lockers and lavatories. This section is usually separated from 
the working area of the floor. 












pers es Se aaa 60 Ft, 0 no nm ie 

AFL OG 2FL Sg SFLO" FLEE or 3FL.O" FLEE 4 FLO" 
oe - a ee 

SMa 

athe ee ——f DOr Ee Bot Ana's PA Dey D3 OS Sng id Tp Ry Oy ES) 

















t 
A 
— =s oe Ha Sema 
WLLL LLZ LLL LLL LLL LL WELTY SS 


7 
/ 


Door Guards: 


ue 
UN at ale ge ch Ft ax 8"X 10! Strut ie 
—8 Ft. 0 f 













= ASS -—-Tread 









ae [<1 —- Riser 


a Main ye 
=eaetg ~y Safety Reinforcement - 


Effective-” 


/ 
Effective-~ Depth 


Thickness 
Details of Stairs 
Fig. 231.—Details of Treads and Risers. (See p. 688.) 


Treads and Risers.—The dimensions of the treads and risers will 
vary with different types of buildings. 

In commercial buildings, the width of tread, exclusive of the 
nosing, measured as shown in Fig. 231, p. 688, may vary from 93 to 


LAYOUT OF STAIRS 689 


10 in. The latter figure should be used in all stairs serving as fire 
exits and also in all stairs that are much used. Secondary stairs, 
such as those leading to the basement, may have 9-in. treads. It : 
a well-established rule that the same width of tread should be used 
throughout the staircase. An exception is sometimes made for 
stairs leading to the second floor, which are used oftener than the 
rest, by giving these a wider tread and a smaller rise, as well as 
greater total width. 

The height of the riser, measured as shown in Fig. 231, p. 688, 
should be as near 7 in. as possible, and should not exceed 72 in. The 
height of the riser is obtained by dividing the distance from finished 
floor to finished floor by the desired number of risers; and the same 
height should be used for all risers, even if it happens to be an odd 
fraction. If the floor heights in a building vary, the heights of risers 
may have to vary, but this variation must be as small as possible. 

Special study should be made of the relation of tread to riser, for 
stairs in constant use, particularly in schoolhouses. If the width of 
the tread is increased, the height of he! riser should be correspond- 
ingly decreased. 

For ease of travel, as well as for appearance, the tread should be 
provided with a nosing, as shown in Fig. 231, p. 688. 

Slope of Subsidiary Factory Stairs—In addition to the regular 
stairs between floors, short stairways are frequently required in the 
operation of a factory. There is always a tendency to make these 
stairs too steep. Not only is this tendency observed in cases where 
it is necessary to avoid occupying too much space or to get in between 
machinery, but one often finds stairways, leading to a balcony or 
platform, so steep as to interfere with rapidity and-convenience of 
travel. This may cause serious loss of time, in case of accident or 
breakdown of the machinery reached by the stairs. 

Where space permits, short flights of stairs of this kind should 
have the same rise and tread as stairs between floors. If this is 
impracticable, one should adopt the rule, based on many years of 
practical experience, on the part of the authors, in factory operation 
and maintenance, that the slope of such subsidiary stairs be limited to 
45 degrees. If this rule is made inflexible, some way will be found 
by the maintenance department to follow it. Stairs steeper than this 
are scarcely, if any, better than a permanent ladder, which occasion- 
ally must be put in. When ladders are unavoidable, spacing of 
rounds should be 12 in. 


690 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


Runs and Platforms.—Runs or platforms, designed to give access 
to shafting or pulleys, or to facilitate the operation of machinery, 
should be built with at least 63 ft. head room to avoid the necessity 
of stooping. If obstructions make it impossible to allow this head 
room, short stairs should be built down and up again into the run, 
since it is less tiring and safer to walk up and down stairs than to 
stoop or bend. 

Intermediate Landing.—If the story height is more than 8 ft., 
the stairs should be provided with an intermediate landing. (Some 
codes allow single flight stairs for story heights up to 12 ft.) The 
width, i.e., the smallest dimension of the landing, should be the 
same as the width of the stairs. In straight runs of stairs (1e., 
when the stairs, the landing, and the next flight of stairs are in line), 
the horizontal distance between risers should be 44 in. 

Live Load.—All stairs, landings, and hallways should be designed 
for a live load of not less than 100 lb. per sq. ft. 


STRUCTURAL DESIGN OF STAIRS 


Two-flight Stairs.—Figure 232 shows the most common arrange- 
ment of stairs in industrial buildings. The stairs between two 
floors consist of two flights with an intermediate landing. ‘They may 
be designed by any one of the three methods described below. In 
each case, an inclined slab is used, with flat lower surface and with 
treads and risers formed on the top of the slab and integral with the 
slab. 

First Method.—The inclined stair and the landing slab are con- 
sidered as one slab, spanning between a beam at the floor level and a 
landing beam placed at the outside edge of the landing. (See Fig. 
232, p. 692.) 

For designing purposes, the span of the slab is assumed to be 
equal to the horizontal distance between the two supporting beams. 
It is then designed as if it were a straight slab. 

The live load is taken at 100 lb. per sq. ft. of the horizontal pro- 
jection of the slab. The unit dead load of the stair portion equals 
the total weight of stairs divided by projected horizontal area. 

Ginee the stairs are usually poured separately from the rest of 
the building and are joined to the floor only by stair-dowels, the 
slab should be considered as simply supported, using M = jul?. 
The thickness of the slab is determined from Formula (11), p. 208. 


STRUCTURAL DESIGN OF STAIRS . 691 


This thickness in the inclined portion is measured on a line at right, 
angles to the bottom surface, from foot of riser to bottom of slab. 
It is marked ¢ in Fig. 232. : 

The reinforcement is placed at. the bottom, longitudinally with 
the slab. Some of the bars are bent up and hooked at the landing 
beam, to prevent cracks due to negative moment. This is an 
important feature, though often neglected. 

Cross reinforcement should be used to prevent temperature 
cracks. As a rule, a 3 bar should be placed in each riser, and three 
or four bars of the same diameter in the landings. 

Landing Beam.—The landing beam, marked B3 in the figure, is 
designed to carry the reaction of the stairs and any wall load that 
may come on it. It may be suspended from the floor beam above, or 
supported by short piers resting on beams below. 

_ In wall-bearing jobs, where the outside walls of the staircase are 
bearing walls, the edge of the landing may be supported on the wall. 
No landing beam is thén required. | 

Second Method.—Beams are placed at both sides of the landing, 
as shown in Fig. 232, p. 692. The inclined stair slab is then designed 
as a slab supported by beams B1 and B2; the span is the horizontal 
distance between beam B1 and the center line of beam B2, and the 
loads are determined as in the first method. ‘The slab may be con- 
sidered as partly restrained at beam B2, which permits the use of 

2 
bending moment o Thickness of the slab in the inclined portion 
is as explained in the first method. 

The reinforcement is placed longitudinally at the bottom. Part 
of the steel may be bent up at beam B2 or special short bars, placed 
near the top surface of the slab, may be used. 

Cross reinforcement should be used as explained above. 

Landing Beams.—Landing beams should be designed as explained 
above, for the proper loads. They may be suspended, or supported 
by short columns. In wall-bearing jobs, beam B2 may be supported 
by the side walls. Beam B3 may be omitted, and the edge of the 
landing slab rested on the end wall. 

_ Third Method.—In this method, the landing beam B2 is retained, 
but B3 is omitted. The inclined slab is then supported by beams 
B1 and B2, while the landing slab acts as a cantilever of the inclined 
slab. 

The slab is designed as follows: The landing slab is considered as 


(02 STAIRWAYS, FIRE EXITS 


lardin ng beam’ 


supported by posts or 
wail or suspended from 
beam above. 


Plan 


Construction Joint 


Stair} 
Dowels 
Main-~ 
Reinforcement 
‘. 


Stair} 
Dowels 


First Method 


Main ---~ 


Reinforcement 


Negative | -< 
Reinforcement : 


Mah) ven hes 
Reinforcement sy 
Joint and Recess” 


Stair 
Dowels 








AND ELEVATOR SHAFTS 


barti tions 


Balance of Panel Flat Slab 
Construction 





Plan- Alternate Design 










Stair 
/ Dowels 


; Main 
Reinforcement 


Negative 
Rein forcement 


Main 
Reinforcement ~ sy F 


Second Method 













rf fee encased 


ncrete 
Cantilever 


Third Method 


Fia. 932.—Structural Details of Stairs. 


(See p. 690.) 


r Reinforcement (. ‘Hooked at Bean) 


7 a eS Sl 
— 


STRUCTURAL DESIGN OF STAIRS 693 


a cantilever loaded by the dead and live loads on the landing and the 
wall load, if any. The load produces a negative bending moment in 
the slab, with a maximum at the edge of beam B2. The inclined 
stair slab is considered as a slab supported by beams B1 and B2. 
The span and the loading are determined in the same manner as for 
the first method. In figuring the bending moment, the moment 
produced by the cantilever should be taken into account, and two 
conditions must be considered: one with cantilever fully loaded, 
which produces maximum negative bending moment; and the other 
with cantilever loaded only by the dead load, which gives maximum 
positive moment. | 

Reinforcement and thickness should be determined for all sections 
of the slab, the thickness of the inclined slab, as in previous cases, 
being measured on a line at right angles to the bottom of the slab. 

The reinforcement should be arranged as follows: Some of the 
bars should run full length at the bottom and extend to the outside 
edges of the cantilever slab. The rest should be bent up at about 
one-fifth of the inclined span and extended near the top to the out- 
side edge of the cantilever slab. If necessary, additional short bars 
should be placed near the top to supply the required amount of steel 
in the cantilever. 

The top reinforcement should be properly supported by blocks 
or chairs. It must be remembered that the cantilever slab relies 
entirely for its strength upon the top reinforcement. Any mis- 
placement of this would materially reduce the capacity of the slab. 
Short rods should extend a sufficient distance from the cantilever 
into the inclined slab to develop full strength of the bar in tension. 

Landing Beam.—The landing beam must be made strong enough 
to carry all the load coming upon it. This will comprise the reac- 
tion of the inclined slab, consisting of one-half of the load upon it; 
the reaction of the cantilever, equal to all the dead and live load; 
and the reaction of the wall (if any). To these should be added the 
reaction produced by the cantilever, which equals the maximum 
bending moment on the cantilever divided by the horizontal span of 
the inclined slab. As in previous cases, the landing beam is either 
suspended or supported on short piers or side wall. 

Three-flight Stairs—Stairs are often constructed with three 
flights, as shown in Fig. 233, p. 694. 

The construction beeen two floors consists of three inclined 
slabs provided with two intermediate landings. 


694 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 























Y /\> A d 1/ ea ecule S 
4h A 2. b -==0 Gs Site A Ad 

ji SIE ae eae aia ot =o NIG, LES ASSIA EGS Ep CLE ~— ] 
uw 


eer gran io oe i ite ~9.4"- * us K 


Plan of Second Floor 


Third Flor = 


Detail of Stair Tread 
and Nosing 













Solid 
Brick Wall,- 






On 


Mm « 
R : 
me w& 
Y 
aes 
—s 
ig ehh 
1 Se 
t 
| 
| 
i 
second Floor § 
ee . Boy 
iat 





Dowels 


Sectionon Line A-A Section on Line B-B 
Frag. 233.—Stairs with Three Flights. (See p. 693.) 





be 
aL. 


ELEVATOR SHAFTS 695 


In the typical design shown in Fig. 233, the stairs are supported 
by a floor beam and by the brick work at the landings. 

The two main sections are considered as slabs simply supported 
by the floor beam and by the wall. The reinforcement should 
extend the whole length of the slab to the edge. As shown in Fig. 
233, in flights leading from the beam to first landing the bars are not 
continuous but are lapped at the juncture of the inclined slab and 
the landing. A continuous bar would have to be bent there in such 
a way that under tension there would have been a tendency of the 
bars straightening by pushing out the concrete in front of them. 

The third section may be considered as spanning between the walls. 
Sometimes the landings have no support on the sides. Then for 
designing purposes, the third section is considered as a simple slab 
supported by the landings. There is no definite point of support. 
The length of the span to be used in computations must therefore 
be determined by judgment. Ordinarily, this may be taken as 
the horizontal projection of the inclined slab plus 2 ft. The bend- 


Pate 
ing moment is found from formula = The reinforcement should 


be extended on the bottom from the outside edge of one landing 
_to the outside edge of the other landing, thus providing for any 
cross bending in the landing. 

The two main sections are considered as rant slabs supported 
by the beam on one end and by the wall on the other end. These 
_ slabs should be designed for the load on the slab plus the reaction of 
the third section. This reaction may be considered as distributed 
over the whole width of the landing. The reinforcement should 
extend the whole length of the slab to the edge of the landing. 

Cross bars should be provided in the treads, as in the other case. 


SPECIAL DESIGNS OF STAIRS 


Stairs may also be constructed by providing beams at the edges 
and then supplying a thin slab between the supporting beams. 
Such construction is seen in Fig. 204, p. 630. 


- ELEVATOR SHAFTS 


The elevators should be placed in groups and where possible 
near the staircase thereby reducing the number of places where the 
continuity of the floor is interrupted. Where staircase is placed 


696 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


outside of the building, as shown in Fig. 280, p. 688, the elevator | 
shafts should also be so placed. 

Elevator shafts always require special framing. To simplify the 
construction, they should be located in such a way as to get the 
simplest framing possible. Sometimes the cost of framing may be 
reduced by introducing additional columns at the corners of the shaft. 
This is particularly advantageous in flat slab construction, because 
it avoids framing of beams into main columns which requires cutting 
of forms for column heads. The supplementary columns, of course, 
must be placed outside of the shaft. 

Sometimes the enclosing concrete walls of the shafts are made 
strong enough to act as bearing walls, thereby saving the cost of 
framing. 

The size of the shaft must be large enough to accommodate the 
elevator car with the clearances required by the manufacturer. The 
shafts must be built with care. All beams must be plumb under. 
each other, otherwise the clear space of the shaft is reduced. To 
prevent accidents, the gate sills should be provided with a nosing 
projecting into the shaft. Usually the sills are also provided with 
safety treads. 

Elevator Pit—A pit at least 3 ft. 6 in. deep must be provided 
below the lowest floor served by the elevator. For all elevators 
serving the basement, the bottom of the pit is below the basement 
slab. Sometimes elevators stop at the first floor, in which case the 
pit may be provided by depressing the first floor a proper distance. 

The pit must have the same clear dimensions as the balance of the 
shaft. ‘Therefore, no part of the footings or columns must project 
into the pit. 

Ordinarily the pits are supported by the ground. The sides and 
- the bottom slab are made 6 in. thick and are properly reinforced to 
resist earth pressure and also to prevent temperature cracks. Where 
the ground is not capable to support the construction the pit walls 
must be carried to the columns or supported on piles. Special 
anchorage is required when there is possibility of upward water — 
pressure. 

When built below water level, the pit should be waterproofed. 

Pent House.—The construction above the elevator shaft is called 
the pent house. This contains the supports from which the elevator 
is suspended and for modern elevators also the elevator machinery. 
The pent house, usually above the roof is properly enclosed and 


= 


ELEVATOR SHAFTS 







oe OS VEG So See SE 
pee (i ES LES Sa ia aoe SON 








~------6+/9"-- ——— 


ee ee! 











Grating 






—----—,f/-'--+-- = <---- 


Section: 


oe ee en ae ---]/'-Q'> Se ee 
we wwe oe ee em rere Hee ---47-/Q"- 


= 











ewewe sore esses oes oc os = 65-9" === wee eme ne eee 










Be —]"-- mm Loe 
Plan 


-o---92tl|"=-- fe. ---------//'-'1------------- 





ae a ere er ere eo eee 


Section 


Fig. 234.—Details of Elevator Shaft. (See p. 698.) 






697 







698 STAIRWAYS, FIRE EXITS AND ELEVATOR SHAFTS 


covered with a sky-light. The size of the pent house depends upon 
the disposition of the machinery. Also there must be enough room 
in case of necessary repairs or replacements. 

The beams in the roof must be made strong enough to carry the 
weight of the pent house walls and roof and also of the elevator, 
counterweight and elevator machinery. These weights are usually 
heavy and must be well taken care of. To take care of the elevator 
loads 100% must be added to the actual loads for impact. 

Grating of proper strength must be provided below the machinery. 

Example of Elevator Shaft Details of elevator shaft are shown 
in Fig. 234, p. 697. This also shows the position of sheave beams and 
the reactions for which the supports must be figured. The loads 
given include an allowance for impact. 


ii ee a a 


CHAPTER XV 
STEEL WINDOW SASH 


Steel window sash is used in the majority of commercial rein- 
forced concrete buildings. The width of concrete pilasters is often 
determined by the width of the -window opening, which in turn is 
governed by the dimensions of the standard window sash. The 
dimensions of the recesses to be provided in the concrete are also 
fixed by the sash. Therefore, to do this work intelligently, the 
designer should familiarize himself with the design of the window 
sash, its standards, and the methods of erection of sash in the build- 
ings. | 

Standards of Window Sash.—The dimensions of sash are stand- © 
ardized as to design and dimensions, and sash of other dimensions 
than the standard can be obtained only at a considerably larger cost. 
Dimensions of windows therefore should be suited to standard sash. 

Steel Sections Used for Window Sash.—Window sash is built of 
specially rolled, solid, low-carbon steel sections. These are strong 
and rigid enough to withstand all the strains to which the sash is 
subjected in handling and in use. The sash frame, however, is not 
strong enough to carry any brickwork above the window— 

The following sections are used in window sash: 

Angle frame members, or outside rails compose the frame of a 
sash unit. 

Muntins are sections placed horizontally and vertically between 
the frame members. 

Weathering members are used at the ventilators to make the sash 
weather-tight. 

Mullions are sections used in connecting sash units to form a com- 
plete sash. | 

Composition of Sash Units.—A steel window sash consists of one 
or more sash units. The rectangular frame for each sash unit is 
built of four angle frame members. Between these, muntins are 
placed horizontally and vertically so as to divide the frame into the 

699 


700 STEEL WINDOW SASH 


required number of lights. A sash unit is often provided with a 
ventilator. Fig. 235, p. 701, shows a sash unit and the cross section 
of the various steel sections composing it, and also the construction 
of the ventilator. 

As evident from the figure, the ventilator is a specially con- 
structed small frame with the top and bottom of regular angle frame 
sections. The sides of the ventilator frame consist of muntins, to 
which, above the pivots, are attached steel weathering members. 
The top and bottom of the opening in the sash unit, to receive the 
ventilator frame, are built of angle frame members, and the sides 
consist of muntins with weathering members attached on the inside 
of the sash below the pivots. The hinges of the ventilator frame 
project beyond the plane of the sash; in some designs the hinges 
form a part of the weathering members. In other designs they are 
part of members that are double riveted to the ventilator and the 
fixed sash. 

The pivot passing through the hinges is placed 2 in. above the 
center of the ventilator opening, to insure proper balance. The 
ventilator is operated by either stay bars or chains. The stay bar 
is attached to the ventilator at the bottom and is provided with 
notches to permit any degree of opening, and when not in use it is 
locked to the frame by a catch. When the ventilator is provided 
with chains, a catch is attached to the bottom member of the venti- 
lator and a pulley to the top, while a clip is attached to the main 
frame to hold the chain. 

Composition of Window Sash—A complete window sash may 
consist of several sash units, all of which are of the same height but 
not necessarily of the same width. The units are connected by 
T-shaped mullions (see Fig. 235, p. 701), each mullion being bolted 
on both sides to the vertical frame members of two adjoining sash 
units. ‘To allow for adjustments in any direction, the holes in the 
mullion, which correspond to the holes in the frame member, are 
oblong in vertical direction, while those in the frame are oblong in 
horizontal direction. The mullion extends at the top to the edge of 
the opening, and at the bottom its flange extends to the bottom of 
the sash member, while its stem extends down 2 in. below the window 
opening. (See Fig. 237, p. 703.) 

Relation of Wall Openings to Sash Dimensions.—As will be 
noted from the details, the sash is secured in the window opening of 
a concrete building by grouting in a part of the flanges of the frame 


RELATION OF WALL OPENINGS TO SASH DIMENSIONS 701 


members and the lower portions of the stem of the mullion. The 
size of the window opening, therefore, differs from the outside dimen- 
sions of the sash; but to avoid confusion, the sash is always desig- 
nated, not by its outside dimensions, but by nominal dimensions 
which correspond directly to the size of the window opening. In a 
window consisting of a single sash unit, the wall opening is identical 
with the nominal dimension of the sash unit. For windows con- 
sisting of multiple sash units, the width of the opening is equal to 
















































a 
uel \ ie 
1 
Ss Ma 
UL 
ie Sy" 
acta fe i 
S Se 
ad ee 
RO 
vox": 
. Se: 
% er 
Loma 90 
C9 G 
w iS oN: 
LS oe: 
SS be 
c \ | ae 
Sere y 











Width of opening PD eect eet 
(For 14% 20" glass ) | 








When desired, stem of 
mullion can be turned in ~ if 
as shown by dotted lines ¢ 


| e sep Sash Unit----=2 i Ay ue Larimer Sash Umtt----- 
2" \ Mullion 


Fig. 235.—Window Sash. (See p. 700.) 


the sum of the nominal widths of all the sash units, plus 2 in. for each 
mullion. In Fig. 235, a horizontal section through a sash illustrates 
the relation between the opening and the sash dimensions. The 
relation of the various parts is evident. 
The relations between the parts of the sash and the total dimen- 
sions are as follows: 
a = distance, center to center, of muntins = width or height of 
glass plus # in.; 
A = distance, center to center, of inside leg of frame members = a 
multiplied by number of lights. 


702 STEEL WINDOW SASH 


Width of sash unit = A + $ in. 

Width of opening = sum of widths of all sash units, plus 2 in. 
multiplied by the number of mullions. 

Height of opening = A + % in. (in figuring A, a is height of 
glass plus 3 in.). 


Recesses for Receiving Sash.—In concrete and brick construc- 
tion, some of the frame members extend into the concrete or brick- 
work and are secured therein by grouting. In concrete, the necessary 
recesses or grooves are formed by nailing, to the formwork of the 
beams and columns, strips of wood of proper dimensions. (See 
Fig. 236b, below.) These strips are removed after the concrete has 
hardened, and form the required recesses in the concrete. In brick 
wall, the frame angles are inserted into a groove between bricks, as 
shown in Fig. 236a, below. For windows consisting of a single unit, 
1 inch reveal is required at the jamb, as shown in Fig. 236¢. 








Ax< Width of Sash nr 5"| Width of Sash 





| and Opening and Opening 
\ gu 
pes: 
a. C 
Jamb for Brickwork dJamb for Concrete Jamb for Concrete 
Multi - Unit Openings , Multi - Unit Openings Single Unit Openings 


Fia. 236.—Details at Jambs. (Se2 p. 702.) 


Erection of Sash—The sash comes to the job in fully assembled 
units. Each unit is placed in position separately, the units next to 
the jamb being sct first. The top flange is moved into the top 
recess, and the side flanges of the frame are moved into the groove 
or recess of the jamb, and the remaining units are then set with their 
mullions. The mullions are first loosely bolted to two adjoining 
frames, the flange of the mullion always being placed outside of the 
frame. The stem of the mullion, in concrete construction, may be 
placed either outside or inside of the sash, as desired. | 

When sills are built in place, the brickwork or concrete of the 
spandrel is erected only to the bottom of the sill. The sash is then 
set and is supported at the bottom on blocks of proper, depth, which 
should be placed at the corners only, to prevent distortion of the 





GLAZING 703 


sash. After all units are assembled, the window is lined up and 
plumbed. The mullion bolts are tightened and the sash is securely 
wedged against the jamb and the head. The sill is finished, and then 
all recesses are grouted. To keep the sash in place, wall ties are 
attached to it and concreted into the sill. (See Fig. 237, p. 703.) 

If stone or pre-cast concrete sills are used, they should be placed 
before the erection of the sash is begun. The sill should be provided 
with horizontal grooves for the stem of the mullions, and also with 
holes for expansion bolts for fastening the sash to the sill. The 
sash is erected from the inside. In setting, the top of the sash is 
first inserted into the groove or recess in the head. The sash is then 





Grout 
} Expansion 
; Bolt 






eight of Sash 


hake " 


nd Opening 


Grout 
(ete ea _-Mullion 


a 


= 
és 
>S 
Q 
~D>> 
<= 
as 
av 
Drs 
= 







eo eee ae - We pee on 
Solid Cut Stone or acs) el eee oe ear Precast— Poured / 

Precast Sills Poured in Place Sills in place 
Back of Sill Poured in Place 





Fig. 237.—Details at Sills. (See p. 703.) 


lifted sufficiently to clear the sill completely, and swing in position. 
Often, the sill is made of two parts, the outside part of stone, with 
the inside part cast in place. The stone sill is placed before the 
erection of the sash is begun. After that, the erection proceeds in 
the same manner as with cast-in-place sills. (See F ig. 237.) 

Painting.—Sash is always painted with one coat at the factory. 
After erection, the sash should receive two additional coats. 

Glazing.—The lights are inserted from the inside after the sash is 
erected. The glass is held in place by wire clips, four per glass. It 
should be carefully back-puttied in such fashion that the glass does 
not touch the steel. Putty is also applied on the inside. Special 
putty is used for steel sash. 


704. 


STEEL WINDOW SASH 


Symmetrical Combinations of Sash 


oe a 


Glass Sizes 


























i peers 18” 14x 20" 
Lights Height Lights Height 
ee pee eh Hite Number 
2 3’ 13” 2 ram wales 9 
3 4! 8” 3 or ae ec 
4 6’ 23" 4 e109” | Wiahre 
5 vie 83" 5 8/ 63” pening 
6 OSS Me 6 LO see 
The over-all widths | The over-all widths 
shown below may be | shown below may be 
had in any of the|had in any of the 
above heights. above heights. 
ay PAM 3/ aes 3 
4! PEYE 4! 102” 4 
5/ aie 6’ 03” 5 
6’ 67 Vf Go 6 
g! 62” 9’ 103” g 
Nee Be 9 
10°F Peo 12h eas” 10 
10) 102” 12° 2G 10 
11 lee 1S di 
12) 41,” pO Re 12 
1S Lip en ash 13 
PESOS Ok i Ci dee add 13 
14) bi Te Sy 14°, 
1502235 LA 14 
16g 0a 18 Gay 15 
Wie wart 197) Vise 16 
19° > 47" 23 ae” 18 
pi Waar a ae Oe 19 
phloem ee 24 ea 20 








Number 
of Sash 
Units 
to Fill 


Opening 


PRoOARRWRWWWWWWNWNNHREE 








Number 
of 
Lights 
in Each 
Unit 





STANDARD SIZES OF SASH 705 


STANDARD SIZES OF SASH 


Glass Sizes.—Standard glass sizes are 12 by 18 in. and 14 by 
20 in. Glass can also be had in 10 by 16 in., and 16 by 22 in. sizes. 
The glass sizes used in ventilators are different from those used in the 
rest of the sash, since the ventilator frame is reduced one inch on 
each side. In ventilators consisting of four lights only, each light is 
reduced by one inch, both in width and in height. For ventilators 
of more than four lights, the outside lights are reduced as above, 
while the inside rows of lights are one inch less in height but the same 
width as standard glass. 

For fireproof buildings, wireglass should be used. 

Standard Sash.—Most manufacturers have standardized the 
dimensions of the sash. The dimensions of the window opening, 
therefore, should be such as to fit the dimensions of the standard 
sash. 

The standard sash is of two types: Warehouse Stock Sash and 
Standard Sash. Warehouse Stock Sash is carried in stock, assembled, 
at the warehouse. Standard Sash, while it consists of standard 
members, must be assembled after the order is received. There is 
ordinarily no difference in price between the two types, but since 
better deliveries can be had on Warehouse Stock Sash, it should be 
used in preference to the non-stock Standard Sash. 

Special sash can be manufactured, but its use should be avoided 
whenever possible, as the price is much higher and a much longer 
time is required for delivery. 

Design of Window Opening.—In designing the window opening, 
select first, the size of glass to be used, preferably from the two 
standards in general use, namely, 12 by 18 in. and 14 by 20 in. The 
larger glass gives the cheaper window. 

Estimate the approximate size of the opening, from the span and 
the desired width of pilaster. 

Decide tentatively on the number of lights and number of sash. 
Then compute the exact size of opening from formula given on p. 702. 
The number of ventilators then should be decided upon. 

The problem may be easily solved by referring to the table on 
p. 704, which gives combinations of different standard sash sizes. 


706 STEEL WINDOW SASH 


UNDERWRITERS’ PIVOTED SASH 


Under some conditions, where it is especially important to make 
the window fireproof, Underwriters’ Sash is used. 

The Underwriters’ Sash differs from ordinary sash, mainly in 
that the glass is held on the inside, in addition to putty, by steel 
glazing angles 4 by 7 in., with the long leg bearing against glass and 
secured to the sash. The ventilators are provided with a chain con- 
taining one or two expansion clips at the top and a fusible link. 
When a fire occurs and the fusible link fuses, the ventilator closes 
and locks automatically. | 


CHAPTER XVI 
STRUCTURAL PLANS FOR BUILDINGS 


A set of plans for a building may be divided into several distinct 
parts. In the present discussion, the architectural and the struc- 
tural plans only are considered. 

The architectural plans consist of floor plans, sections, and ele- 
vations. They give an idea of how the complete structure will look. 
Usually, the structural features are only indicated. The dimensions 
of the structural members are given only when they affect the archi- 
tectural features or when, for any reason, they are limited as to size. 

The structural plans show the size and arrangement of the struc- 
tural members. The portions of the construction which do not con- 
tribute towards its strength, and which are built after the structural 
frame is erected, are usually omitted. Thus the piping,.the location 
of partitions, unless built of concrete—brickwork in skeleton struc- 
tures, and similar features are not shown on the structural plans. 
Inserts also are shown separately. 

Structural plans are of two classes: 

1. General plans, forming a part of the architect’s set, 

2. Detail working plans to be used in the field. 


GENERAL STRUCTURAL PLANS 


General structural plans are prepared before the contract is 
awarded and form a part of the contract drawings. The com- 
pleteness of these plans varies with the practice in the offices of the 
various engineers and architects. 

1. Complete Structural Plans—Some firms prepare complete 
structural plans, giving the concrete dimensions and the required 
reinforcement for all members in the structure. 

2. Typical Design Only—In many instances, the general struc- 
tural plans give only the design of typical members. The specifica- 
tions give the live loads and the allowable stresses. The actual 

707 


708 STRUCTURAL PLANS FOR BUILDINGS 


design of all members is left to the contractor’s engineer, with the 
understanding that the methods outlined in the design of typical 
members will be followed and that the plans are subject to the 
approval of the architect’s engineer. 

2 Floor Construction not Designed—Often in flat slab construc- 
tion, the footings, columns and beams are either fully designed as in 
Case 1, or indicated by typical designs as in Case 2; while the selec- 
tion of the type of floor construction or system of flat slab reinforcing 
is left to the contractor, with the understanding that the selected 
design must be acceptable to the architect or his engineer. The 
live load, stresses, and bending moments are specified. 

4. No Structural Plans Shown.—In a large number of instances 
(the majority of buildings), the architectural plans submitted for 
bids do not include any structural plans. The live loads are specified. 
Sometimes specifications also have requirements as to the stresses 
and bending moments. In cities having building codes, this is 
omitted and the specifications call for a design which would satisfy 
the requirements of the building code. The structural design is left 
to the contractor. | 

The method to be adopted in any particular case depends upon 
circumstances. 

Theoretically the first method, that of preparing complete struc- 
tural plans, should give best results. Actually, however, this method 
is recommended only where the organization preparing the general 
plans is composed of engineers well trained both in theory and field 
practice of reinforced concrete construction. | 

Under ordinary circumstances the engineers of architects’ offices 
cannot specialize in concrete construction, but must deal with all 
structural materials as necessity arises. While such experience 
tends to broad knowledge of structural engineering in general, it 
precludes the complete mastery of details of reinforced concrete. 
Therefore, either the second or the third method will give much 
more satisfactory results, especially when the working plans are 
checked. 

The method where no structural plans are shown, is not recom- 
mended unless the working plans are checked by a competent engi- 
neer employed by the architect or unless the working plans are pre- 
pared by an engineer selected by the architect. If the matter is left 
entirely to the contractor, the safety of the structure depends alto- 
gether upon him or upon the engineer engaged by him. It is obvious 


GENERAL STRUCTURAL PLANS 709 


that this method either through intent or negligence may lead to 
unsafe designs. It is absolutely essential for the safety of the con- 
struction that the architect or owner employ an engineer in whom 
they have confidence, rather than to rely altogether upon the engi- 
neer of the material man or contractor. Often reliance is placed by 
architects upon the fact that the designs are checked and approved 
by the Building Department. In practice only the largest cities 
maintain engineers skilled in concrete construction. Even there 
mostly the checking is not complete but is confined to main features. 
In small cities the approval of plans is perfunctory. 

Purpose of General Structural Plans.—The purpose of the gen- 
eral structural plans is: 

1. To enable the contractor to estimate the required amount of 
concrete, steel, and formwork. 

2. ‘To serve as a positive guide in preparing the working drawings. 
These objects are not attained unless the plans and specifications 
show clearly what is wanted, particularly as regards the points upon 
which there is a possibility of a difference in opinion. If the engi- 
neer preparing the general plans does not make his intentions clear, 
the design, in these details, is left to the engineer preparing the 
working plans. . 

General Instructions.—Kach office should establish rules govern- 
ing the method of preparing structural drawings. To make the 
drawing clear, the lines having different meanings should be of 
different thickness and style. The following rules are suggested: 

Building line, dash and dot line, dashes about 2 in. long, medium 
thickness, diluted ink. 

Center lines of columns, light long dash and dot lines, diluted ink, 
dashes over one inch long. 

Dimension lines, light solid lines, diluted ink. 

Outline of concrete, solid black lines, medium thickness if visible, 
dash line of same thickness if invisible. 

Reinforcement, heavy solid lines in elevation. In section use 
solid circles for round bars, solid squares for square bars. 

Sections of concrete members should either be shown as concrete 
in the conventional way or shaded on the back of the tracing with a 
soft pencil. Care should be taken that the shading is not so heavy 
as to obscure any lettering nor lines of reinforcement. 

Objection may be raised to showing reinforcement by solid lines. 
It is usually invisible in the concrete, and dash lines would be more 


710 STRUCTURAL PLANS FOR BUILDINGS 


logical. However, (1) dash lines require much more time to draw, 
and (2) in many cases when bars meet at an angle, dash lines do not 
make clear just where each bar goes. Solid lines, therefore, are 
preferable and should always be used. 


COMPLETE GENERAL STRUCTURAL PLANS 


Complete general structural plans consist of foundation plan and 
footing schedule; floor and roof framing plans with girder, beam, 
and slab schedules; and column schedules. The architectural 
plans usually show sufficient general sections and elevations, so that 
these are not repeated in structural plans. Special partial sections, 
however, will be found necessary. 

Numbering of Columns and Footings.—Two methods of number- 
ing columns and footings are now in general use. 

In one method, the panel pomts on the floor plan are numbered by 
consecutive numbers, 1, 2, 3, 4, etc. The numbers are then used to 
designate the columns and footings at these points, and so Cl, C2, C3, 
C4 would indicate columns 1, 2, 3,4 and FI, F2, F3, F4, etc., corre- 
sponding footings. These designations are used in column and footing 
schedules. This method is simple, but it has the drawback that it is 
often difficult to locate the columns on the plan, especially in a large 
or irregular building. Also, duplication of numbers is possible, 
especially when new columns are added after the plans are completed. 

In the second method, which is preferred by the authors, the 
mark of the column is fixed by its location. This is accomplished 
by the use of coordinates in the following manner. The main 
column lines that are horizontal on the plan are designated by small 
letters. The top column line is called a, and the others, in alpha- 
betical order, b, c, d, e, ete. The main column lines that are vertical 
on the plan are designated by numerals, 1, 2, 3, beginning at the left. 
The panel points, located at the intersections of the two sets of lines, 
are designated by the letter and the number belonging to the inter- 
secting lines. These designations are used for columns and footings. 
For instance, the panel point at the intersection of line b with line 3 
is called b3, the column located there is called C—b3, and the footing 
F-b3. | 

Special column points located between the main column lines 
are designated by the addition of 4 to the number or a prime sign to 


COMPLETE GENERAL STRUCTURAL PLANS “11 


the letter. Thus, panel point b3% is located on line b between lines 
3 and 4. b’3 is located on line 3 between lines b and ec. 

This method of designation fixes the position in the building of 
every panel point. It is, therefore, preferable to the first method. 
Some contractors claim that the use of a letter and number is con- 
fusing in the field, but unquestionably the advantages overbalance 
any possible disadvantage. 

Numbering of Beams and Girders.—Typical beams and girders 
are designated by a number indicating the floor, a letter, B, G, or L, 
designating a beam, a girder, or a lintel respectively, and a second 
number indicating the distinctive beam or girder on the floor. Thus 
2B3 would mean beam 3 on second floor. On one floor, there are | 
usually a large number of beams of the same span with the same 
concrete size and amount of steel. All these beams are given one 
common designating mark, which is marked on the floor plan in each 
place where the beam occurs. The dimensions of the beam are 
given in the schedule. It is advantageous to maintain the same 
designating marks on all floors, so that a beam 3B3 in the third floor 
will correspond to a beam 2B3 in the second floor and 5B3 in the 
fifth floor. Where the framing in the various floors is sufficiently 
different, this latter plan is not workable, and beams should be 
numbered on each floor without regard to numbers in other floors. 

Where the panel points are designated by coordinates, as explained 
in the previous paragraph, the odd beams and girders, which occur 
only in one place, may be designated by their location. .For a hori- 
zontal beam, the panel point at its left extremity is used; and for a 
vertical beam the panel point at its top end. Thus, a beam on the 
second floor, starting from the intersection of lines d, and 4, may be 
called 2B-d4. In a.building having vertical beams, the location 
may be still more definitely fixed by adding h for horizontal beams 
and v for vertical beam. Thus, 2B—d4v is a beam in the second floor, 
starting from panel point d4 and running vertically down. For 
intermediate beams, the prime sign and + may be used, as explained 
in connection with designation of columns. For illustration of this 
method, see Fig. 239, p. 714. 

This method of designating beams has been adopted as standard 
by several large contracting firms. It is of special advantage in the 
drafting room. If all beams are numbered consecutively, on the 
other hand, and in the process of design some odd beams are elimi- 
nated, the remaining beams must be renumbered or else the sequence 


712 STRUCTURAL PLANS FOR BUILDINGS 


is destroyed. Also, beams are sometimes added after the schedule 
is finished, so that, for instance, beam B45 may be in the same panel 


with beam B3. This makes the checking difficult. 


FOOTING SCHEDULE (For Stepped Footing) 


st Sten | 2nd. Step], ]_Reinforcemert poner 
¥ B edesta emarksS 
eres: 


C 
it — 
6-6"| 420" | 5" B6'Sq 20-2'9 |22-¢'9 | 
Gd A NN NUNN 











































Qa ae 
Stepped Footings Sloped Footings 
Fic. 238.—Method of Showing Independent Footings. (See p. 713.) 


Foundation Plan.—The foundation plan, usually drawn to a 
scale 1 in. = 1 ft., shows the location of the footings and basement 
walls. The elevation of the basement slab may also be shown on the 
foundation plans. This is especially desirable where parts of the 


basement are lower than the rest. 
When the condition of the ground is known, the elevation of the 


COMPLETE GENERAL STRUCTURAL PLANS 713 


footings should be given. This helps in estimating the amount of 
excavation and the length of the pedestals. 

The outline of the footings may be shown on the plan. The 
rest of the information necessary for independent footings may be 
given in a schedule. A good form for a footing schedule is given 
below. ‘This in conjunction with the sketch, gives all information 
required for independent footings. It is evident that the same 
schedule may be used for square and rectangular footings. 

There is more difficulty in showing combined footings. Schedules 
will be found insufficient in this case. Details of combined footings 
should be shown in a sketch, which should give the concrete dimen- 
sions and the required steel. The reinforcement should be clearly 
shown by elevations and sections, in the position in which it is to be 
placed. This is important, as the design of combined footings is not 
sufficiently standardized. It is obvious that it would not be suffi- 
cient to give the number of bars only, without showing their location. 
If there are several combined footings of the same type, one sketch 
may be used for all. The various dimensions may be indicated by 
letters, and a schedule made up in the same manner as for independent 
footings. The length of the bars should be shown in the sketch in 
terms of the span. The determination of the actual length of bars 
may be left for working drawings. | 

The design of the wall should be shown by sections, which should 
be provided wherever there is a decided change in the height of the 
wall or the type of reinforcement. The reinforcement should be 
shown in such a way as to make the intentions of the designer clear. 
If horizontal bent bars extending into adjoining spans are used, one 
typical bending sketch at least should be shown; or a note should 
state clearly the design desired. 

Floor Plans.—Floor plans may be drawn to a scale 4 in. = 1 ft. 
and in some cases to a scale 7; in. = 1 ft. The larger scale is pre- 
ferred where the floor arrangement is at all complicated. 

Where the arrangement of beams does not vary from floor to 
floor, one floor plan may be used for all typical floors, even if the live 
loads are not the same. The difference in size of beams or in rein- 
forcement is taken care of by the schedules. 

For any floor in which the bulk of the construction is typical and 
special arrangement is restricted to a small portion of the floor, the 
typical floor plan may be used in conjunction with a partial plan of 
the special section. The partial plan should be properly tied to the 


714 STRUCTURAL PLANS FOR BUILDINGS 










J " 
Hee 


By 
Ee 


opg2H 2BetH 2B4 284 OB4 86a 




















































> 
bio 
Pe oe) 
eo 
red i<c) 
[so] a 
a 
= 
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a Ps} Ss = a °) 
Zo =o =e Tey 
aQ ol 
s ee 
is —— stat 1)- 
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im a = Lm Se a lS 
2Ba5HL|L | | L #1 Je a| ae-@® 
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am TS ol eal col aor wate pase] spas [ouch] TS) 
22611 13164 SET 13-6 5413-6 18-62 13-6 oo Oo 13:6— to 4-19410!"<— 17.6" 240-6"" @ 
2 
2 
\ " 
= 13% 6 Bars per bay 
& ~3 713-6 Je 13-6 
Se 73019" SECTION ‘'AA” eee eae 
! 16 
= 1410" =S 6" 








2Ba5H 
15. 
iF 





DETAIL OF BARS AT_“‘AA'* 


m 


Column Schedule 
olumn Numbers ree 













p3|— 














ov 

Be E 6 to 
os SI a g13 
AS a 






















DHT 18-07 9- FT 4-3" 
2-3") 18-0'°1-5°F 4’-0" 
4-8") 15’-0” ic 


































13 |opi1’x19 18-2’T 07-3” 

T 197-0"| 8-8’T 19-0” 

ge r 21-§. 4-1’T 10’-0"| 21-2" 6'-3” 
a 8 20-2°T 8'-9"|21-2°T 5'-6” 10-8'T 4’-3” 





14-¥T §-6 
1-5/T 5/-6" 
§ 


Ms 
Af tol 
8-3"T 17’-3 


Usual Column Sections Not Shown 
2 





2B2 


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oBhI iH 
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| | 


66 60 © © © GO 6, one 





< 2B2 
ote roe tite eee LL Da ht , 
9.4 4-1 8- “6 -6—7— 13-6 3-6 3-6 
9-4 13: aes 13 13-6 1 1 1 


Note: For cross-section through this building see Fig. 290, p. 808. 
Sections through facades are shown in Fig. 265, p. 754. 


Fig. 239.—Typical Framing Plan and Steel Schedule. (See p. 711.) 
Buildings 1, 3 and 5, Massachusetts Institute of Technology, Cambridge, Mass. 
Stone & Webster, Builders. Sanford E. Thompson, Consulting Engineer. 





COMPLETE GENERAL STRUCTURAL PLANS le 


typical plan and a note should be inserted on the typical plan, 
referring to the special arrangement. To prevent confusion, many 
engineers prefer to give an extra floor plan where any change in beam 
arrangement occurs. 

A beam, on 3-in. scale plans, is indicated by two lines; and on 
qe-in. scale plans, by a single heavy line. The latter scale should 
not be used unless the position of the beams is clearly shown else- 
where. When the architectural plans give sufficient information as 
to sizes of openings and locations of beams, this information does 
not need to be repeated. 

The roof plan is treated in the same manner as a floor plan. In 
many cases it is possible to use the typical floor plan for the 
roof. . 

Beam and Girder Schedule.—The location of the beams and 
girders are shown on the floor plan. Their dimensions and rein- 
forcement are best shown in a schedule. 

A typical beam and girder schedule is shown below. The best 
method of showing concrete dimensions is to give the width of stem 
and the total depth of beam or girder. Thus, beam 12 in. x 24 in. 
is a beam the width of stem of which is 12 in. and the total depth 
24 in. A note should state whether depth is measured from finished 
surface or rough concrete. 

Since the schedule does not always give the length of bars, it is 
essential for estimating purposes to show clearly how far the bars 
should be carried into the column or the adjoining span. In case 
hooks are used, their dimensions should also be given. In connection 
with the straight tension bars, it should be stated whether or not they 
are intended as compression steel at the support. If so, they must be 
carried far enough into the support to develop the compression 
stress, and this determines their length. Information regarding 
the lengths of bars and their arrangement is best covered by sketch 
of typical interior and exterior span. The length of bars may be 
shown in terms of the span, as in Fig. 240. A general note should 
state the conditions under which the design is made. The note 
should give the distance from bottom to outside face of bar. It 
should state that the same amount of steel should be placed at the 
support as at the center. All special requirements should be stated, 
so as to make it possible to estimate the amount of steel, with suffi- 
cient exactness, and later, to enable the contractor’s engineer to make 
complete working drawings. 


716 STRUCTURAL PLANS FOR BUILDINGS © 


Method of Showing Details.—Details of beams should consist 
of elevations and section as shown in Fig. 240, p. 716. The bending 
of the bars should be shown to scale in the elevation. Bars should 
pe labeled in elevation. This information should not be repeated. 
When straight and bent-up bars are used, for the sake of clearness 
the bars are drawn in the elevations in several layers, even if in actual 
construction they are intended to be placed in one layer. The actual 
position of the bars should be shown in the section. If no section 1s 


BEAM & GIRDER SCHEDULE 


SpecialBars | __Stirryps | 
‘ : N2& Size, i 
N2&Size|Location |“sksicn’| Pach ena |e 




























1B) | 12x24|2g $ \23 F 










er 
TN el Sere oe ee ee 


— 
a 





/" : : 
6h g 2 Stirrup Ties 





‘Ming Diam, for Sq. Bars 
oe n Rd” 


Fic. 240.—Beam and Girder Schedule. (See p. 716.) 


used, it should be stated in a note that bars are shown in several 
layers for the sake of clearness and that actually they should be 
placed in one layer. 

Where reinforcement is more complicated, it may be shown in 
elevation in the position occupied in actual construction. The bars 
are then drawn separately so as to give the bending of each bar 
clearly. This method 1s used in folding Figs. 214 and 215, for show- 
ing reinforcement for frames. It is evident that it would be im- 
possible to make the arrangement of steel clear in any other way. 


COMPLETE GENERAL STRUCTURAL PLANS fou 


Slab Details and Schedule.—The thickness of slab and the 
required reinforcement are often shown directly on the floor plan. 
If these vary for different parts of the slab, a special slab schedule 
may be used. The customary mark for slab is 5, and the method 
used is that described for beams. 281, 282, 253, for instance, 
would indicate different parts of the slab in the second floor. Infor- 
mation concerning the arrangement of slab steel at the support is 
often neglected, leaving an open field for mistakes in estimating. 
The purpose of the general drawings is to enable the estimator to 
make his estimate accurately, and the necessity for guesswork on 
his part must be entirely eliminated. It is evident that if no design 
is prepared the estimator is at liberty to use his own standard 
practice, which may differ from the assumption of the engineer and 
thus result in friction between the engineer and contractor. In 
case designs are made but essential details are omitted, the con- 
tractor must protect himself by assuming excess material to allow 
for differences of opinion, and this will result in increased cost. 
To avoid excess estimates, the designer preparing general designs 
must make his requirements absolutely clear. For details of slab 
reinforcement see p. 210 and folding page opposite p. 579. 

Flat Slab Details and Schedules.—The flat slab reinforcement 
details should be shown by a floor plan and.a schedule. 

The thickness of the slab should be marked on the floor plan. 
The dimensions of the drop panels, column heads, and _ brackets 
should be dimensioned on the floor plan and also shown by typical 
sections. The reinforcement should be divided into groups of bars. 
Each group should be given a mark consisting of a letter and a number, 
the letter designating the ‘group and the number the particular 
member of this group. The schedule should give the reinforcement 
for each group of bars. The same designating marks, of course, 
are used in the schedule and on the plan. 

Reinforcement of a typical interior and exterior panel should be 
given in detail. These should show clearly how the bars should be 
bent, how far they should extend into the support, and how far they 
should extend into the adjoining panel. The lengths should be given 
in terms of the span, so that the details may also serve for odd 
panels. 

The floor plan in Fig. 241, p. 718, the schedule and the bending 
sketches show the method clearly. The same method is used both 
for the four-way and two-way systems. In addition the plan should 


718 STRUCTURAL PLANS FOR BUILDINGS 








Flat slab schedule Ist to 4th floor. | { 
Slab Bin 1” grano. finish, bonded eH | 
column head 4-3" mameter Drop panel “i \ 
4" x 6'=10"x 6-10" Concrete 1-2 4 Leiaserptle boot -laor 






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400%; 
bed — — 9-0" — ahem = 19’-0'--<-— = 19-05 =| 
Nore. Provide supports for a// bent bars consis ting of 
chairs or conerete blocks of proper height 


Fic. 241.—Partial Plan and Schedule for Flat Slab Building. Two-way System. 
(See p. 717.) 







—— ey 


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; Flat Slab Schedule /st. to 4th f/oor 
Slab &" I" Granolthie Jinish, bonded. Concrete 1.2/4 
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ile 10°-0"bia. | 172 9"Dia. 
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NOTE Sketches for bars not shown They are similar to those 
BIs> gale in ae ee Page 369. 7 






















“H 
Hf 
es 


Otss 


sera Te 





” 


-0 





—-19 


aa 


“Thi 


No 


——-19'-0 
GY 


> 
i ed 


Fhe -19% Ot eo tig: pee 0"-—> 


Fig. 242.—Partial Plan sag and Sections f 








Section /-/ 









Section 2-2 


—Brick 








Section 
#Slab building. Smulski System. 


fea 


oie! 


(See p. 719.) 


(To face page 719.) 












i ty . <a “ Ee] 
ye Gouase . 
: ae ae . 

‘ ‘ ; : 





£2344 aon sant of) Rs BA 





COMPLETE GENERAL STRUCTURAL PLANS 719 


show details of typical interior and exterior panels as shown in 
Figs. 126 and 127. 

In the Smulski (S. M. I.) system, the reinforcement is divided into 
units. The units between columns are designated by letter A, the 










C> to Cg inal 
D2 to Dg ind. 
£,to£gind 












Col load in 000% 
Column Size 
Core S/ze 








oS S12 
; Ny la 
+ 52 yy. A . 


3 












Vertical Bars 
Spirals or Ties 





























.| 298 

S a | 36°20" 

in | CoreSize [| 32%/6" ki 
Ver tica/Bars|~ 2" . 


Spirals or Ties 











‘Core Size 
Vertical Bars 



















Bas. Spirals or Tres 
S _| Dowels 
to Pedesta/ 
Top of | 
Footing 56" 
A 





Warrable 






Notes:_ Give thickness required for fireproofing 
Give length of spirals in respect to heights 
Give required lap for vertical bars A 
Give any other information reguired by special conditions 


Fig. 243.—Column schedule. (See p. 720.) 


center units by letter B, column head units by letter C, and bars 
across unit A by letter T. A number placed before the letter may be 
used to designate the story in which the unit is used, while a number 
following the letter serves to differentiate between varying units of 
the same type. (See Fig. 242, opp. p. 719.) 


720 STRUCTURAL PLANS FOR BUILDINGS 


Column Schedule.—The column schedule should give concrete. 
dimensions, concrete mix, number and sizes of vertical bars, spacing, 
and size of column ties. For spiral column it should give diameter 
of core, size of wire, and pitch of spiral. The height of column also 
should be given in the schedule, as it assists the estimator in getting 
his steel tonnage and is an aid in preparing general working drawings. 
The load for which the column is designed should also be given. 

Norss should give all the other information not contained in the 
schedule. | 

The thickness of concrete for fireproofing should be clearly shown 
on a sketch or covered by a note. It is necessary to state how much 
lapping is required for column steel, and how high the spiral should 
be carried. A sketch of the column ties should be shown, particu- 
larly in cases of unusual columns. \ 


WORKING PLANS 


The difference between general structural plans and working 
drawings is evident from comparison of the purposes the two sets of 
plans are intended to serve. The eeneral structural plans are 
‘ntended for making up estimates and as a guide for preparation of 
final working drawings. The working plans are intended for use in 
the field. They must be complete and exact. Dimensions of all 
concrete members must be given. Not only the number and size of 
reinforcing bars, but also their length must be given. 

Recesses and Dowels.—It is important to show on the working 
plans all recesses, dowels, and nailing blocks required in the concrete 
work. Recesses are required, for example, for window sash, while 
at the beams in staircases recesses and dowels should be shown. 
When concrete walls are used, dowels may be required in floor slabs 
to connect the wall to the slab, and also pour holes through which 
the wall concrete may be poured. Nailing blocks are required to 
nail the flashing and the doors, and sometimes at .the windows. 
The importance of providing recesses and nailing blocks is evident, 
as drilling holes in solid concrete adds unnecessarily to the cost. 
Some contractors do not show any recesses On the drawings, but 
rely upon the superintendent or carpenter foreman to make proper 
provisions. This practice is not satisfactory. 

Inserts.—Special plans should be made for the location of inserts. 
An attempt to show them on the floor plan, with other structural 


—— 


WORKING PLANS | 721 


data, would result in confusion. Usually, details of typical plans 
are sufficient. | 

Building Lines.—To facilitate laying out the building in the field, 
the plans should be tied to a set of lines called “ building lines.’ 
These dre straight lines for each face of the building, which usually 
coincide with the main outside face of the building. For rectangular 
buildings, the building lines are rectangular; for polygonal buildings, 
they are polygonal, and so on. 

The building lines are established on the ground, and all measure- 
ments are made from them. All recesses in the wall, for instance, 
are measured from the building line. The outside faces of the con- 
crete for beams and columns are also located from the building 
line. 

The building lines should be shown on the floor plans and in 
sections through the wall and wall beams. In sections, the building 
line is indicated by a vertical line. The distance from outside face 
of concrete to this line should be given. For instance, in a brick 
veneered building the building line would coincide with face of brick. 
The concrete work is then 5 in. from the building line. 

Foundation and Footing Plan—The information regarding 
numbering columns and footings, on p. 710, applies to working plans 
as well as to general plans, and all should correspond. 

The foundation plan should have all the information required 
for the laying out, in the field, of all footings and basement walls. 
Footings may be shown in the manner described in connection with 
general plans (see p. 712). Outside footings should be tied to the 
building lines. 

Footing schedules should contain, in addition to the information 
given on general plans (see p. 712), the number of footings of each 
type and the number and length of bars in each footing. For com- 
bined footings, bending sketches, completely dimensioned, should be 
given for each type of bent bar. The position of all bars in the 
footing should be definitely fixed, in a clear manner, either by a 
detail drawing or a note. The column dowels should also be given 
in the footing schedule, rather than in the column schedule, because 
the steel is required at the time the footings are built. The necse- 
sary information as to the spacing of dowels should be given. In 
spiral columns, this is fixed by the diameter of the column core. 

The basement wall should be shown by sections and, if necessary, 
by elevations. Sections should be taken, not only for typical rein- 


722 STRUCTURAL PLANS FOR BUILDINGS 


forcement, but also where the height, thickness, or amount of rein- 
forcement differs from the typical. The section should show height 
of wall, thickness, reinforcement, distance from building line, and 
all other information required in the field. Not only the size and 
the spacing of the bars should be given, but also their length and 
their number. The position of the first bar should be fixed. If 
bars are bent, a bending sketch should be given. The position of 
the bent bar should be shown in elevation. This is of particular 
importance where the bars are placed in an unsymmetrical position ; 
thus, in end panels, it is necessary to show how far into the column 
the bars should extend. 

Sometimes the different panels of the wall are given designating 
marks, and each panel is marked. A schedule, similar to that used 
for slabs, gives the required ‘nformation as to thickness of wall and 
amount of reinforcement. 

Elevations of the basement slab should be given. Sections should 
be given where the elevation of basement wall changes, showing the 
retaining wall. All pits should be clearly indicated, with dimensions 
and their locations tied either to a building line or a column line. 
Cross sections through pits should show depth, thickness of wall, 
and details of reinforcement. 

Sizes of all openings in the wall should be clearly given, and 
openings should be located with respect to column lines and floor 
elevations. 

Floor Plans.—Floor plans should be drawn to a scale } in. = 1 ft. 
If floor construction is complicated, a larger scale should be used. 
Floor plans show the position of beams, the slab reinforcement, and 
sections. 

Building lines should be shown on the floor plan. Where the 
center line of beams does not coincide with the center line of columns, 
the offset must be clearly shown. This is best accomplished by 
sections. Not only typical sections through the spandrel beam 
should be shown, but also sections in all places where the concrete 
dimensions or the distance from building line differ from the typical. 
All openings in the floor must be clearly dimensioned and should be 
tied to the adjoining center lines of columns or to building lines. 
When the floor plan is too complicated, especially if it shows slab 
reinforcement, the concrete dimensions and the details of the open- 
ings should be shown on a separate plan or a partial plan. This 
applies particularly to stair and elevator openings, which are usually 


WORKING PLANS 723 


shown to }-in. scale in connection with stair details. Where recesses 
occur, they should be indicated in plan or in section. 

Beam Details.—All beams should be marked on the floor plan 
by the method explained on p. 711 in connection with general plans. 
Beam schedules should give complete information. A convenient 
schedule, more complete than the one suggested for general structural 
plans, gives the following data: number of beams; more complete 
information on concrete dimensions; not only the depth and width 
of the beam, but also the depth of the slab on each side. If the 
section of the beam is irregular, so that full information cannot be 
conveyed by the schedule, a cross section through the beam should 
be shown. 
| The information on reinforcement, also, is more complete than 

for general plans. Not only the number of bars but also their length 
is given in the schedules. To insure proper location of the bars in a 
beam, a detail of the beam with full reinforcement, including stirrups, 
should be given (see Fig. 240, p. 716). This is particularly necessary 
where the reinforcement is not symmetrical. In the end beams, 
this should show how far the hooked ends of the bars must extend 
into the column. Some designers give this information in connec- 
tion with bending sketches, by locating the ends of the bars with 
respect to the centerlines of columns (see Fig. 245). Marks of bent 
bars and stirrups (as noted below) should be noted in schedule. 

Straight Bars in a beam are not provided with any mark, as on the 
job all straight bars of equal size and length are stacked together, 
irrespective of their destination in the structure. This simplifies the 
work. | 

In listing bars, the straight bars are kept together, arranged by 
sizes beginning with the largest, and also grouped by length. 

Bent Bars should be given a designating mark, and each bar 
after it is bent should be tagged with its mark. Aluminum tags 
should be used and securely tied to the bars to prevent their loss. 
This is especially important where bent bars are shipped any distance 
and require repeated rehandling before they reach the job. Paper 
and cloth tags, sometimes used, are not satisfactory as they often 
get lost. This not only creates confusion on the job, but also is the 
cause of bars being used in the wrong position with great detriment 
to the construction. A bar without tag must be measured, and the 
bending sketch thus made compared with the bending schedules, to 
determine where the bar belengs. This is expensive in time and 


724 STRUCTURAL PLANS FOR BUILDINGS 


money. Both the engineer responsible for the job and the con- 
tractor should insist upon secure tagging of bent bars. | 
Marking of Bent Bars.—Whatever method is used for marking 
bent bars, any possible duplication of numbers must be prevented. 
A simple method is to give the bent bars the mark of the beam. 
Thus, if a beam is marked, in the schedule, 2B3, the bent bars may 
be also marked 2B3. If more than one type of bent bar is used in a 
beam, the different types may be differentiated by adding a small 
letter to the beam mark. Thus, 2B3a, 2B3b, and 2Bé8e are three 
types of bent bars in beam 2B3. This system of marking is not 
adapted to stirrups, as often the same type of stirrup can be used for 
a number of beams of different mark. Marks U1, U2, etc., may be 
used for all stirrups of the same dimensions, irrespective of their 
location. Thus, if beams B1, B3, B5, and B7 have the same con- 
crete dimensions, they require the same type of stirrup and in each 
case the stirrup is marked Ul. Sometimes, to prevent confusion in 
marking, a number designating the floor is prefixed to the mark. 
In such cases, 2U1 is a stirrup for a beam in the second story. 
Another system of marking which is sometimes preferred, but 
which does not indicate location of bar and is thus less definite, 
employs figures only. , 
By this plan, the mark of each bar consists of three figures, such 
as 725. The first figure designates the size of the bar, representing 
the number of eighths of an inch in the diameter or side of the bar. 
Thus, a 4-in. bar will be represented by 100;. a 4-in. bar by 200; a 
3_in. bar by 300; a g-in. bar by 400; a 5_in. bar by 500; a f-in. bar 
by 600; a g-in. bar by 700; a l-in. bar by 800; a 14-in. bar by 900, 
and so on. Bars of the same diameter but of different length and 
bending are differentiated by the last two numbers, which are con- 
secutive. ‘The first bent bar in the 4-in. diameter list will be 701, the 
next 702, and so on. This method was used in Fig. 215, opp. p. 664. 
Bending Sketches.—Every bent bar must have a detail sketch. 
It is not sufficient to show the general outline of the bar, as in general 
structural plans. All dimensions must be carefully worked out. The 
importance of this cannot be over emphasized. Poorly bent bars, 
even if detected before they are placed, must be rebent at the job at 
a higher cost than that of the original bending and often with vexing 
delays. The required dimensions are shown in Figs. 244 and 245. 
In double bent bars the important dimensions for bending are 
the height of the bend, measured from outside to outside of bar 


WORKING PLANS 725 


(usually called “out to out ’’) and the horizontal length of the bend. 
The inclined length of the bent is not used in the field, but it must be 
computed to determine the length of the straight portions of the bar. 
If shown on bending sketch, it assists greatly in checking. 

Bending sketches are usually made separately from the beam 
schedules. In the beam schedule, the mark of the bar is given and the 
type of bending indicated by a small sketch, which is of assistance 
in locating the bar when needed in the field. In the steel schedule, 
the same mark is used. 

For convenience, bars of the same type of bending are placed on 
the same sheet. Thus, bending sketches are made for groups of 
double bent bars, groups of rings, and groups of stirrups. 


BENDING:.-SKETCHES 








For trussed bars, separate bending sketches are often drawn for 
each mark of bar. This method requires more work in the office, 
but it simplifies the work in the field. The foreman can give a 
workman a sketch representing each bar to be bent by him. 

Separate bending sketches may also be used for the purpose of 
indicating the position of the bars in the beam. Thus in Fig. 245, 
p. 726, the position of the ends of the bar is located in respect to 
the enter line of the columns. This is important where the position 
of the bar is not symmetrical. | 

Another method is to use one sketch for a number of bars, as 
shown in Fig. 244, p. 725. The sketch is drawn at the top of the 
schedule, and the dimensions for each bar are given in the schedule 
under the different headings. By proper marking of dimensions, 
one sketch can be used for all trussed bars listed on the sheet, with 
and without hooks. In Fig. 244 one sketch was used for four types 
of bent bars. This simplifies the work in the office. It is evident, 


726 STRUCTURAL PLANS FOR BUILDINGS 


however, that the understanding of this schedule requires higher 


intelligence on the part of the workman than that of the other plan. — 


Stirrups may be shown either by making a drawing for each stir- 
rup or by making a table similar to the one suggested for bent bars. 

Slab Details in Beam and Slab Construction.—In many cases, all 
the information about slabs may be given on the floor plan. Thick- 
ness of slab, elevation, and type of finish should be marked directly 
on the plan. Reinforcement may also be shown on the plan unless 
the arrangement of panels is complicated, in which case a slab 
schedule must also be prepared. Sometimes the main reinforcement 


fecpoutside face of column | 


aN 
Tay Lees cae 
18"! Out to Out a 
















Le of column 


aye ear mee Te _ oe 20'- Ole = ee 


chose oh oat 
ae 
| Out to Out x 





20" 






Fic. 245.—Separate Bending Sketches. (See p. 725.) 


is shown on the floor plan, while odd panels are covered by a schedule. 
Not only the spacing of the bars but also their number and length 
should be given. The position of the first bar in each section is 
located by a dimension from some fixed line. The remaining bars 
are located sufficiently by giving their spacing. Method of bending 
should be clearly shown. In many designs, bars of one panel length 
are used. Straight bars alternate with bent bars. The first straight 
and the first bent bars are located, and the rest are covered by giving 
the number of bars of each type and the spacing. Where bars extend 
over several panels, the bars of one type of bending alternate with 
bars of some other type of bending. Sometimes bars of the same 


f 
Peek os. « 


WORKING PLANS (27 


type of bending are used but the ends are staggered. Each type of 
bar should be located as explained above. All these expedients are 
used to supply a sufficient amount of steel, not only in the center but 
also at the support, and with the smallest tonnage practicable. It is 
important to make the intention of the designer clear, since mis- 
placement of bars may give an excess of steel in one section and not 
enough steel in some other section. The more complicated the 
arrangement, the more clearly it must be shown. 

Position of the bars in the slab should be shown by a section 
indicating the type of bending required. When bars are to be bent 
kefore placing in the form, proper bending sketches should be pro- 
vided. The straight bars are then listed separately from the bent 
bars. Where the bars are “ hickied,” i.e., bent after they are placed 
on the form, the location of the points of bending should be given, 
and the height of the bend should be worked out and shown. In 
this case, all bars are listed as straight bars. 

Flat Slab Construction—Working drawings for flat slab con- 
struction should be worked out in the manner described on p. 717 
in connection with general plans. The steel schedules, however, 
must be complete, and the length of all bars must be given. In the 
schedule for the Smulski System the length of each ring and the 
required length of lap must be given. 

If the bars are bent before placing, bending sketches must be 
prepared in the same manner as for beams. Different types of bars 
should be given an appropriate mark on the schedule. Bars should 
be tagged with an aluminum tag. If steel is bent on the forms, all 
bars should be listed as straight bars, care being taken to allow for 
the proper length required for the bend. 

Column Schedules.—Column schedules for working drawings 
differ from those described on p. 720 only in that the lengths of all 
bars are given. The number of ties and their bending sketches are 
also given. Where the column bars are bent at the ends on account 
of difference in size of the upper columns, bending sketches must be 
prepared. ; 


CHAPTER XVII 


ARCHITECTURAL TREATMENT OF EXTERIOR AND 
INTERIOR OF REINFORCED CONCRETE BUILDINGS! 


By Henry C. ROBBINS 


In this chapter the architectural treatment of the exterior and 
interior of reinforced concrete buildings is discussed. Illustrations 
of exterior and interior treatment, as well as details of construction 
of different types of buildings, are given. 

The development of the exterior treatment of reinforced concrete 
buildings has grown quite naturally with the increase in their use 
and with the growing knowledge of the architectural possibilities 
of this type of construction, which was originally developed for 
utilitarian purposes rather than for artistic merit. 

Beginning with the kind of building which frankly expressed its 
structural materials in piers, girders, and curtain walls of concrete, 
with liberal window openings, the concrete building has passed 
through various stages of architectural expression to its present 
state, in which, to all outward appearances, it may differ in no 
essential feature from other equally substantial forms of building 
construction. 

The clothing of the reinforced concrete building with brick, stone, 
or terra-cotta, or a combination of these materials with concrete, 
makes possible a variety of architectural treatments which in the 
details of construction present no more serious difficulties than are 
ordinarily met in the design of buildings of other types. The interior 
finish of concrete buildings may also be as simple or as elaborate as 
the requirements demand, without seriously .taxing the ingenuity 
of the designer as far as the practical execution,is concerned. 

It is intended in this chapter to describe and illustrate some of 
the accepted methods used in common practice to produce satis- 
factory results in both exterior and interior finish of such buildings. 

1 The authors are indebted to Henry C. Robbins for this chapter, which has 
been especially prepared by him for this treatise. 

728 


a 


EXTERIOR WORK ¢29 


EXTERIOR WORK 


For the purposes of this discussion, the exterior treatment of 
concrete buildings may be divided into: 


(1) Concrete exterior; 

(2) Concrete in combination with brick, tile, or other masonry ; 
(3) Brick exterior; 

(4) Stone or terra-cotta exterior. 


CoNCRETE EXTERIOR 


Simple Treatment of Concrete Exterior.—The simplest form of 
exterior treatment is found in the frank expression of the concrete 
- structural members with curtain and parapet walls of the same 
material, as illustrated in Fig. 246, p. 729. Buildings of this type 
are generally found in the warehouse or factory building class. 








Fic. 246.—Simplest Form of Factory Building: Concrete Piers, Concrete 
Spandrels, and Curtain Walls. (See p. 729.) 
Turner Construction Co., Builders. 


Details of concrete exterior are clearly shown in the part elevation, 
Fig. 247, and section, Fig. 248. As is evident from the illustrations, 
the exterior columns form the piers between the windows. The 
width of piers is governed, not only by the magnitude of the loads to 
‘be supported, but by architectural requirements and by the desired 


720 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


width of the windows. With steel sash, the window extends from 
pier to pier, therefore the width of piers is governed by the standard 
width of sash. When more than one window in a panel is required, 
different window units may be separated by concrete mullions. The 
piers are joined together by spandrel beams at the floor levels, acting 
as lintels of the windows, and also as supports for the floor construc- 
tion. Above the spandrel beams, and filling the space between the 
floor and the bottom of the windows, are concrete curtain walls. In 


Parapet Concer. Cap-------- x 
Flashing--..... 
Roofing me 3 






















Cinder Bire 























Sash 
Grouted in--" 
























































Sasi 
Grouted in~--. 


SIM Cast a 
Separate--"| 5] 


Curtain Wall.-- 
































































































































Reinforcement not Shown --- 





— sash Grouted Section 
ee ————— through 
= Wall 





Detail below Window Sl aS te 
Showing fe: —— 


_ Curtain ‘all -~ 


Plan of Pier between Sash 
Fic. 247.-—Details of Concrete Exterior. (See p. 729.) 


the simplest form, their tops may be simply beveled to receive the 
wood window frame. Very often, especially when the steel sash is 
used, a separate sill is cast on top of the curtain wall after the sash 
is in place. 

The roof spandrel beam may be continued above the roof level 
to form a parapet with a beveled top, or with a simple, square-edged 
projecting cap to give a finish to the wall. 

The simplest and cheapest arrangement of exterior wall treatment 
is that in which the face of the piers coincides with the face of the 
spandrels. This, however, presents a rather crude and monotonous 
appearance, 


EXTERIOR WORK 731 


The monotony may be relieved at very slight additional expense 
by setting the spandrel beams back from the face of the wall columns. 
The curtain walls are then placed slightly back from the face of 
the spandrel beams, and the concrete window sill is made to project 
over the curtain wall. Such an arrangement is shown in Fig. 248, 
p. 731. The appearance is also improved by the introduction of 
stronger or more massive treatment of end panels. This is fre- 
quently justified by placing the stair towers in these locations and 























- Wood Sill 






































Wood Sill- - 
Concrete Sill -- Reinforcement of — Concrete Sill 
Curtain Wall Not 
Shown 
Concrete __ __-Concrete 
Curtain Wall Curtain Wall 
eas Face of Column 
_.. Face of 
Column 
_-. Concrete 
/ Spandrel Beam ». 
yi 
_-7 Anchorage Strip~.__ 
“for Wood Frame ~~~. ° 
; fs = =o ee : 
Window Head Rea Ne - Window Head 
lial viet 
sd gis 
Concrete Curtain Wall Set Concrete Curtain Wall Brought 
Back From Face of Spandrel Out Near Face of Spandrel 
s Showing Concrete Sill Ouverhanging 
Inside to Take Wood Window Frame 
b 


Vic. 248.—Sections through Spandrel Beam. Concrete Exterior. (See p. 731.) 


changing the windows from the typical arrangement followed in the 
rest of the building to a smaller size with surrounding structural 
members of increased width. 

The reveals and projections obtained by the methods described 
above help to break the monotony of an otherwise flat surface by 
the casting of shadows, which, after all, are responsible for much 
of the effect in all architectural composition. 

The concrete curtain walls are shown in Fig. 248a and 6. The 
construction in Fig. 248b may be used where the thickness of the 
curtain wall is less than the thickness of the window frame, 


732 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


The concrete curtain walls are usually built separately after the 
frame is constructed. ‘To prevent temperature cracks, they must 
be properly reinforced. (See p. 384.) Also, they must be tied to 
the building, and recesses or slots must be left in the columns to 
receive these walls as they are cast. 

The treatment of the surface of the concrete itself is discussed 
under a separate heading below. (See p. 734.) 

Elaborate Treatment of Concrete Exterior.—It is possible to carry 
the exterior design of concrete structures into quite elaborate detail, 
and very pleasing effects may thus be obtained. The means used 
for the purpose are: (1) more elaborate details of molded work; 
(2) special treatment of concrete surfaces; (3) combination of the 
two methods. 

The selection depends upon the purpose for which the building 
is to serve and upon the extra expense that can be devoted to orna- 
mentation. Pleasing effects may be obtained at a comparatively 
small cost, a combination of simple molded work with surface treat- 
ment of concrete. 

More Elaborate Detail of Molded Work.—The appearance of con- 
crete buildings may be improved by the addition of molded belt or 
string courses and cornices, and in more expensive buildings by the 
‘ntroduction of special ornamentation. In selecting the design, the 
character of the material must be kept in mind. The molding and 
the design of the cornices should be such as present the least 
problem in form construction. Projecting members with square 
corners, simple molded forms such as quarter rounds or plain cove 
moldings with flat fillets, plain sinkages, and other similar forms are 
the easiest to produce and are best adapted for concrete work. 
The effect should be obtained by general proportions of the mass 
rather than by refinement of details. 

A simple but effective treatment of concrete exterior is shown 
in Fig. 249, p. 733, where the building is provided with simple cornice, 
pilaster caps, and bases. The entrance received rather more elab- 
orate treatment. 

Molded details may be poured with the structure, or they may 
be precast and placed in the construction in the same manner as 
other stone trim. 

When molded in place, the moldings can be poured simultaneously 
with the rest of the construction, or the supporting frame may be 
built first and the belt courses and cornices poured separately. 


EXTERIOR WORK 733 


The latter method simplifies the formwork, as the forms for the 
projecting moldings are supported on the structure instead of on the 
elaborate support which would be necessary if the molding were 
poured simultaneously with the structure. In this way the molding 
ean be built truer to line, and better workmanship can be obtained. 
Also, special concrete may be employed. When belt courses are 
poured separately, proper recesses must be left in the supporting 











Fic. 249.—Building of the Frank E. Davis Fish Company, Gloucester, Mass. 
(See p. 732.) 
Densmore, LeClear & Robbins, Architects and Engineers. 


structure, while steel dowels, often called “‘ bonds,” should be pro- 
vided to tie the two parts of the structure. To prevent cracking, the 
belt course must be reinforced by separate horizontal bars. 
Wherever molded work is cast in place, it should be arranged, if 
possible, to bring the end of a day’s pouring at the molded section, 
thus concealing the line that is always in evidence, in the finished work 
of plain surfaces, where one day’s work ends and the next begins. 
Important decorative work is best done in concrete pre-cast in the 
form of “ cast stone ” and built into the main structure. See Fig. 250, 


724 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


p. 734, in which the cartouches on either side of the entrance and the 
key-block were cast separately and inserted. 

Treatment of Concrete Surfaces.—Much can be done to produce 
an interesting exterior by the choice of surface treatment, which, 














Fic. 250.—Entrance Detail, Carter’s Ink Factory, Cambridge, Mass. (See p. 733.) 
Densmore, LeClear & Robbins, Architects and Engineers. 


if carefully handled, goes far toward making up for the lack of 
elaborate molded work. Exposed concrete may be finished in a 
variety of methods, ranging from the smoothing of the surface, 
primarily to eliminate form marks and to fill voids, to the picking or 
hammering of the surface to reveal an aggregate chosen particularly 
for its color. 


EXTERIOR WORK 735 


A dense, smooth surface is desirable for concrete which is to be 
exposed to view and upon which no great amount of finishing work 
is to be expended. To produce such a surface, thorough spading 
of the mass as it is deposited is essential, and tapping the 
outside of the formwork with a hammer or mallet is helpful 
in further settling the mixture into place, thus forcing the heavy 
aggregate back from the surface’and bringing the mortar to the 
face. | : 
For all concrete work which is to be exposed to view, it is desir- 
able that the formwork be tight and smooth. It should be erected 
true to line and supported in a substantial and workmanlike manner. 
The materials should be as free as possible from defects which 
would show in the surface of the concrete. 

In the usual commercial work, however, it is impossible to 
eliminate all voids, irregularities, and fins or slight projecting ridges 
which are caused by the mortar oozing through the cracks in the 
formwork. ‘These defects must be corrected if a reasonably regular 
and unbroken surface is to be obtained. 

Surface Treatment by Rubbing.—Most satisfactory results are 
obtained at the least cost when the finishing is undertaken while the 
concrete is still green. The fins then can be smoothed down and 
the entire surface brought to an even finish, by rubbing with a 
wood or cork float, using a mortar composed of one part of cement 
to two parts of fairly coarse sand as an abrasive, the mortar at the 
same time filling the voids. 

With concrete which has set hard, it is obvious that more work 
is involved in producing a satisfactory finish. The fins must be 
removed by hammering, chipping, or rubbing with a carborundum 
stone, and the voids must be filled. After this has been done, a 
mortar composed of one part of cement to one part of screened sand 
is rubbed in, by means of a cork float manipulated in a rotary manner. 
The surface should be thoroughly saturated with water before the 
mortar is applied. 

After the mortar, applied as described above, has thoroughly 
hardened, the entire surface is rubbed with a carborundum stone 
while water is being applied with a brush, until a smooth, even 
surface, uniform in tone, is obtained. This method was used in 
finishing the building shown in Fig. 249, p. 733. 

It is essential that the mortar be thoroughly rubbed in during the 
first operation, so that no appreciable thickness of skim coat remains. 


736 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


Otherwise, the coating will crack and peel off, if exposed to extremes 
of temperature. 

An effective variation in the treatment described above may be 
obtained by using white cement in combination with the fine sand, 
in place of ordinary gray cement. In this case, the second rubbing, 
after the mortar has set, may be dispensed with. When this is done, 
the thin surface coating should be kept wet as it sets. 

The so-called spatter or dash coats, so frequently employed in the 
finish of domestic stucco work, are not commonly used in reinforced 
concrete buildings of the commercial or industrial type, as they 





Fic. 251.—Example of Brush Finish. (See po 100) 


usually involve expense in the preparation and finishing of the 
surfaces to be treated, out of proportion to the results to be obtained. 

Brush Finish.—Green cement may be effectively treated also by 
washing with water to expose the aggregate, using a wire brush or a 
coarse, stiff fiber brush. Good judgment as to the proper degree of 
hardness of the concrete is required in using this method, since if 
the surface is too hard, the brush will not be effective in removing 
the mortar, while if it is too green, particles of aggregate will be 
dislodged, leaving a coarsely pock-marked surface. Ordinarily, 
this method of finishing must be undertaken in about twenty-four 
hours from the time of pouring. (See Fig. 251, p. 736.) 





EXTERIOR WORK 137 


Bush Hammer Finish.—A more effective and also more costly 
finish than those described above is obtained by hammering or 
picking the surface to expose the aggregate. The bush hammer 
commonly used in stone masonry is a satisfactory tool for this 
purpose. 

The brushing method previously described approaches the 
hammered or picked treatment in result, but it lacks the sparkle of 
the latter because hammering not only exposes the aggregate, but 
cuts it as well. 

The condition of the concrete for a satisfactory hammered treat- 
ment must be the reverse of its condition for brushing; that is, it is 
necessary for it to be thoroughly set and hard, so that the aggregate 

is imbedded firmly enough not only to prevent dislodgment as the — 
- mortar is being cut back, but to withstand the actual cutting of the 
aggregate particles. This method presents a difficulty in the 
finishing at external corners, where it is practically impossible to 
prevent the cutting out of pieces of aggregate, thus leaving a ragged 
arris or corner. This may be obviated by first rubbing the surfaces at 
the corners to a smooth finish, as described above, and then hammer- 
ing up to a sufficient distance from the corners to leave a smooth draft 
line, of a width to be in scale with the hammered field. | (See Fig. 252, 
p. 738.) This treatment may be elaborated by choosing the surface 
aggregate for its color, but the common run of stone or gravel used 
throughout the work ordinarily will provide a satisfactory surface 
when exposed and cut. 

Coloring of Surface by Selection of Aggregates.—If a special color 
of aggregate is desired and is available only in small quantity or at 
an extra cost, it may be deposited in the outer surface only af the 
portions to be so treated. ‘This may be done by inserting a sheet- 
metal dam about 3 inches back of the inner face of the formwork, 
and depositing the special mixture between the dam and the form- 
work at the same time that the regular mixture is being poured in 
the balance of the member. The dam is raised gradually with the 
progress of deposition of the concrete. The spading is confined to 
the space above the bottom of the dam, else the two mixes may 
run together. It is not practicable to introduce this method in 
columns and piers, where the reinforcing steel, and particularly the 
column ties, complicate the handling of the dam; but in spandrel 
beams and walls, or in unreinforced members, it is quite feasible. 

Another method of introducing special concrete near the surface 


738 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


of the building was used in the construction of the Hide and Leather 
Building in New York City 

After the stecl reinforcements for the spandrel had been fabricated 
as usual, expanded metal was wired to the bars, leaving a distance 
varying from 13 to 2 in. between the expanded metal and the outer 





Fic. 252.—Example of Hammered Surface Surrounded by Smooth Draft Lines. 
(See p. 737.) 


form. Into this outer space a special surface concrete was poured, 

consisting of one part of white Portland cement and two parts of a 

colored aggregate, the latter being composed of green stone chips, 

feldspar and quartz, ranging in size from 7 to 1 in. This surface 

concrete, which was hand-mixed and of dry consistency, was guided 

into the front space by a broad, flat chute, and tamped in place with 
2Thompson & Binger, Architects and -Contractors.. 


EXTERIOR WORK 739 


a special T-head as the latter being made by fastening a 
1-in. flat, 6-in. long, to a 4-in. bar about 12 ft. in length. 

The balands of the beam, i.e., the space between the expanded 
metal and the rear form, was aired with the usual concrete. Pre- 
cautions were taken to pee the facing concrete pile up about 1 ft. 
higher than the structural concrete during the process, to prevent 
the common aggregate from flowing to the outside face. After the 
forms for the exterior surface had been removed and the necessary 
patching done, the concrete was permitted to season during the 
winter; in the early spring it was ready to receive its final finish. 
The two lower floors were bush-hammered by stone masons. The 
upper floors, sixteen in number, were surfaced by a carborundum 
_ grinding Pasieting. after which a colorless waterproofing was applied 
over the entire exterior. This treatment was given mainly to prevent 
the elements from attacking the surface of the concrete. 


CoNCRETE IN CoMBINATION wiTH Brick, TILE on OTHER Masonry 


This combination may be used in warehouse and factory’ con- 
struction, where the curtain walls are built of ordinary hard brick, 
and the cost approaches the cost of the simple all-concrete type. 
With proper treatment of concrete surfaces and use of more expensive 
filling materials, very ornamental effects may be obtained. 

Combination of Brick Curtain Wall with Concrete Piers and 
Spandrels.—The most common construction of brick and concrete 
is that in which the piers and spandrel beams are of concrete and 
the curtain wall of brick. Such construction is shown in Fig.. 253, 
p. 740, where the curtain walls and parapet are _of red brick and the 
rest of the fagade is carried out in concrete. ‘The section in Fig. 
254 shows the construction more fully. Asin the case of the all 
concrete exterior, the size of the concrete pier is mainly governed 
by the required architectural proportions and the width of the 
window sash. The exposed concrete beam, which in the illustration 
rests entirely below the slab, carries the weight of the curtain 
wall and windows and also any floor load that may come upon it. 
The construction of the sill on the top of the brick curtain wall 
depends upon conditions and particularly upon the type of window. 
With steel window sash, concrete sills are commonly used, and are 
built after the sash is in place. 





} BUILDINGS 


CONCRETE B 


L TREATMENT OF 


ECTURA 


740 ARCHIT 









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EXTERIOR WORK 741 


An interesting variation from the ordinary treatment of spandrel 
beam and curtain wall, when the beam is entirely below the level 
of the floor which it supports, is the “ upstanding ” spandrel, which 
involves a somewhat less efficient structural design for this particular 
member, but has the advantage of 
bringing the tops of the windows close 
to the ceiling. Figure 255, p.. 742, 
illustrates this form of construction 
and may be compared with Fig. 254, 
p. 741, the latter showing the usual 
dropped beam. 

The brickwork in the spandrels 
- may be plain or elaborate, the elabo- erick 
ration consisting of the use of differ- 
ent grades and colors of brick or 
different patterns. 

Brick or tile work is often intro- 
duced into concrete piers, walls, and 
spandrels in the form of pattern work, 
making possible most effective treat- 
ments in both color and design. The 
possibilities in the combination of 
these materials are limited only by the : 
ingenuity of the designer and the re- pig. 254 Seetion fProughic pare 
strictions of expense. It is, of course, drel Beam. Concrete and Brick 
obvious that wherever concrete ap- Exterior. (See p. 739.) 
pears it may be finished in any 
of the methods discussed under “‘ treatment of concrete surfaces.” 

The brickwork in the curtain wall may be replaced by ornamental 
terra cotta. Figure 256, p. 743, shows an interesting use of archi- 
tectural terra cotta in curtain walls and in the spaces over the 
arches just below the cornice. The spandrels are elaborated by the 
use of terra-cotta pediments at the third-story level. The terra-cotta 
work is green-glazed and gives a most striking appearance in combina- 
tion with the concrete, which is finished in white cement. 

Consideration of the general proportions of the mass and con- 
centration on a dominating feature bring about the same satisfactory 
results in architectural expression as may be obtained with the same 
study in any other type of construction. Figure 257, p. 744, is an 
excellent example of the possibilities of such study. While the main 


Wood Sill 





~~ Masonry Sill 


- Brick 


-- Air Space 


: aan — ne ‘ 
a -Window Head 


742 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


body of the building is given a touch of color in the brick spandrels and 
parapet wall, the tower is carried out entirely in concrete cast in place. 

Concrete Piers, Brick-faced Spandrel, and Brick Curtain Wall.— 
The concrete spandrel beam, which is exposed in the previously 
described type, may be cov- 
ered with brick. The wall 
construction then consists 
of exposed concrete columns 
with the balance, except for 
sFace of Column the windows, covered with 
brick. 

Figure 258, p. 745, illus- 
trates this method. In this 
case, as in all cases where 
masonry is to be placed 
against concrete as an ex- 
Anchorage Strip terior facing, a space of about 
For Wood Frame one inch should be allowed 
between the face of the con- 
crete and the back of the 


Fic. 255.—Upstanding Spandrel Beam Per- tse to allow for varia- 
mitting Window Head to Come up Near tions in the surface of the 
Ceiling. (See p. 741.) concrete member. The con- 


crete beam is then set back 
from the outside surface of the finished wall, a distance equal to the 
width of brick plus one inch. 

When masonry is to be used as a facing for concrete, it is cus- 
tomary to support the facing material on structural steel angles 
securely bolted to the concrete members. For ordinary cases, 
3.in. bolts spaced 24 in. apart are used. This may have to be. 
increased where the angles carry heavy load. The bolts, with heads 
and washers, or with the ends bent for anchorage in the concrete, are 
inserted in the formwork, with a threaded end left projecting a suf- 
ficient distance to pass through the leg of the angle and receive a nut 
and washer on the outside. These bolts may be held tight, during 
pouring, by nuts, one just inside and one outside of the formwork. 
The steel angle supporting the brick facing is usually 6 in. deep in 
order to get the anchor bolt well up into spandrel beam. 

The spading of the concrete where the bolts occur must be done 
with care, so as not to displace the bolts or throw them out of line. 











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Masonry Sill -- 223 






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EXTERIOR WORK 745 


Since the nature of concrete formwork is such that precision 
in the spacing of anchor bolts is somewhat difficult as compared with 
shop work in structural steel, it is well to allow for some slight varia- 
tion, particularly in the horizontal direction of the bolt spacings, by 
slotting the steel shape instead of punching it. (See Fig. 258, p. 745.) 

It is not practical to slot the angle vertically, but some allowance 
may be made for inaccuracy in vertical dimension, by making the 
slot slightly wider 
than the diameter 
of the bolt. This 








sell) - Wood Sill 
requires the use of r f Gis rer 
asonry Sill - Y) 3 
a heavy washer un- LA |: 
Wi 
der the nut and a Yj 
thorough tightening puscpace-- Lop 


“Face of Column 





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of the latter to pre- 
vent slipping under Brich-- 
the load of masonry. 
Backing of Cur- 
tain Walls.—It is Concrete eet 
obvious that the : 
backing -of curtain 
walls, described sieropnk 
above, may be of : 
the same material 
as for similar work 
in other types of 
fireproof construc- Frc. 258.—Spandrel Beam Faced with Brick, with Brick 
tion. The problem Curtain Wall. (See p. 742.) 
will vary with the 
character of the building and with the degree to which it is 
desired to avoid condensation. In most warehouses and _ fac- 
tories, a small amount of condensation is not harmful. In such 
cases the curtain walls are made from 6 to 8 in. thick, without any 
special precautions. In structures such as office buildings, schools, 
hospitals, and buildings for habitation, where moisture inside must 
be avoided as far as possible, the choice of the backing material and 
the method of laying it should be given prticular attention, to prevent 
the actual driving through of rain and also to eliminate condensation. 
Water troubles frequently appear even after the utmost care has been 
used in building, but the following methods may be used as being 


-) Sometimes Made of 


This Portion of Wall Is 
Single Course Brick 


-Fill Up With Mortar to Here 


Possible Irregularity 
in Face of Spandrel 


- Brick Facing 
-Angle Iron Support 


——————— 


Anchorage Strip. et 
For Wood Frame 


\ Angle Slotted 


Window Head - For Anchor Bolt 


746 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


good practice without involving great expense and as giving reasonably 
satisfactory results. 

If the wall is to be of brick, an air space of 1 or 2 in. in width 
should be provided between the outer and inner courses, and care 
should be taken that this space be free from mortar. 

A further precaution is to coat the inner surface of the outer 
coursing with a waterproofing material, this being done either as the 

wall is built up or after 
the outer coursing 1s built 
up to level and before the 
inner section is started. 
eieacgh er The outside face of the 
wall may be coated with 
a colorless waterproofing 
agent. The use of either 
of these methods of applied 
Face of Column waterproofing does not 
affect the conditions as re- 









Wood Sill-— 


Interlocking Tile 
Smooth Face 





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Brick Facing effective insulation. The 
backing of the wall may 
be of hollow tile of the in- 
Sh LRU terlocking variety, which 
provides for bonding with 


Anchor -—~~_ 


Anchorage Strip - as 
for Wood Frame |.2° 37 


a 


Fic. 259.—Spandrel Beam Faced with Brick, | . 
with Curtain Wall of Face Brick and Interlock- brickwork, or of the ace 
ing Tile. (See p. 746.) mon form of hollow tile 


bonded with brick or with 
metal ties. Figure 258, p. 742, shows a wall with brick backing, while 
Fig. 259, p. 746, shows a wall with backing of interlocking tile. 
For buildings where the interior faces of these curtain walls are 
not to be plastered, the methods of construction described above 
are within the limits of good practice, since the amount of moisture 
which will come through is not of serious moment and is, under 
ordinary circumstances, evaporated before it becomes troublesome. 
When walls are to be plastered, however, it is the opinion of the author 
that no method of construction or treatment is as satisfactory or 
positive as the furring of the inside of the wall to provide vertical 










— Anale Iron Support 













EXTERIOR WORK 747 


air spaces. This may be done with ordinary wooden or stcel furring, 
upon which is placed the steel lathing and the plastering, or with 
tile furring blocks laid with the hollow spaces vertical. 


Brick EXTERIOR 


The next development beyond the partial use of masonry in 
connection with concrete is the complete clothing of the concrete 
skeleton structure with brick, in much the same manner as is prac- 
ticed in buildings of skeleton steel construction. Figure 260, p. 748, 
shows a group of buildings entirely covered with brick and with 
no indication as to the type of construction. 

This illustration was selected to show that concrete Wane may 
be used irrespective of the elaborateness of the architectural treat- 
ment. 

A less elaborate construction is shown in Fig. 262, p. 750. In 
the buildings shown in both illustrations, the outside treatment is 
independent of the size of concrete wall columns. These are, there- 
fore, designed only for the load, as their size is not governed by 
architectural considerations. ‘There the outside face of the columns 
must be in-one vertical plane, any reduction in size of the structural 
column is obtained by setting back the inside face in the different 
stories. If the exterior design calls for a buttressed column, the 
structural member may be reduced in thickness, as the column goes 
up, to correspond to the changes in the set-backs of the exterior 
face. 

In factory and warehouse construction, where the window occu- 
pies a large part of the wall, an arrangement shown in Fig. 261 may 
be used. ‘The window sash extends from concrete pier to concrete 
pier. The veneer is applied outside of the building only. The 
shape of the concrete pier is as shown in the figures. The recesses 
in the columns are easily obtained by using rectangular forms with 
blocking of proper dimensions inserted in the corners. 

The concrete wall beams ordinarily are set back just enough to 
allow a veneer of brick equal in thickness to one width of brick 
(usually 4 in.) plus one inch allowance for variations in the face of 
the concrete member. This may vary as necessary to meet the 
architectural treatment. If possible, the spandrel beam should be 
made deep enough to reach to the top of the window. ‘This saves 
the expense of extra lintels over the windows, 





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EXTERIOR WORK 749 










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Supporting angle sl Ls to Aca! there 


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Detail Elevation 
Details _of Brick Exteror 




























































































































































































Masonry Ties” ain Mas asonry Ties \ Brick Facing 


Details of _Facing_of Columns with Brick 


Fig. 261.—Partial Elevation and Sections of Brick Veneer Building. 
(See p. 747.) 


TREATMENT OF CONCRETE BUILDINGS 


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EXTERIOR WORK | 751 


Support of Veneer.—The support of brick facing or any other 
such architectural treatment on the face of the structural members 
is accomplished by angle supports, in the same manner as described 
above in the discussion of masonry spandrel walls. (See p. 742.) 
The angle supports should extend along the spandrel beam and, in 
case of wall columns, across the face of the columns also. If span- 
drels are flush with wall columns, continuous angles may be used. 
When the spandrels are set back, separate short angles should be 
used across the column. The spacing of bolts and the method of 
securing them to the forms is the same as given on p. 742. 

The steel supports for the facing should be bolted onto the 
structural members at each story. The angle on the spandrel is 
usually 6 in. deep to get the anchor bolt well into spandrel beam, 
while the angle on the column face can be 4 in. deep unless stronger 
angle is required by the load. 

Ties for Facing.—In addition to the angle support, metal ties 
should be used for anchorage of the masonry to the concrete. These 
should be placed vertically, not more than six courses of brick apart. 
When the spandrel beam is no more than 20 in. deep, no ties are 
required. Horizontally the eae may be 24 in. In the simplest 
forms, the ties consist of 4-in. wire, preferably brass or copper, 
imbedded partly in the concrete and extending from it a sufficient 
length to reach well into the brick masonry. In constructing, holes 
are drilled in the form, the required distance apart, and the wire 
inserted to be buried in the concrete. The part of the wire to be 
imbedded in brick extends outside of the form. 

To avoid the expense of drilling holes, ties of special design may 
be used. With these, the part intended to serve as anchorage in 
masonry is tacked to the formwork, while. the part to be imbedded 
in concrete is bent back at right angles to the formwork. After the 
concrete is poured, this part of the tie becomes buried in the concrete 
mass. When the forms are stripped, the ties will detach themselves 
from the forms, but will of course remain anchored in the concrete. 
The ties should be of sufficient length and of such flexibility that the 
portion left outside the concrete surface can be bent to meet the 
coursing in the masonry and be laid into the joints as the facing is 
built up. If this method is used, it is not necessary accurately to 
predict the masonry jointing or to place the ties exactly at joint lines. 


752 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 











Fic. 263.—Building of the Salada Tea Co., Boston, Mass. (See p. 753.) 
Densmore, LeClear & Robbins, Architects and Engineers. 


EXTERIOR WORK 


753 


STONE OR TERRA CoTtTa EXTERIOR 


The use of stone or architectural terra cotta for the facing of 
reinforced concrete structures ordinarily presents no more difficult 
problems than are met in similar buildings where steel forms the 


structural members. 











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Showing Stone Facing 


Fic. 264.—Sections Showing Facing 
with Architectural Terra Cotta and 
Stone. (See p. 755.) 


Figure 263, p. 752, shows a concrete building, the lower stories of 
which are faced with limestone and the upper story with architectural 


terra cotta. 


In general, the manner of support and anchorage of 


the stone or the terra cotta is similar to the methods described above 


in the discussion of brick-faced buildings. 


In the building illus- 


trated, the first two stories of stonework rest on the foundation wall, 


754 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 






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INTERIOR FINISH 759 


while the remainder of the facing is carried on angle supports at each 
floor level. 

Particular attention must be given to the anchorage of members 
having sufficient projection beyond the normal face of the wall to 
introduce possibilities of overturning. The use of heavy ties an- 
chored in the concrete and turned into the stone, or of angle shapes 
secured to the structure and overlapping the top of the stone at the 
back, are methods frequently used; but each case is worthy of 
special study to determine the most satisfactory method. See 
Fig. 264, p. 753, for an illustration of facing with stone and archi- 
tectural terra cotta. 

Sections through ornamental facades on major and minor courts 
of the Massachusetts Institute of Technology buildings,? are shown 
in Fig. 265, p. 754. Attention is called to the method of facing 
of the concrete and also to the different methods used for anchoring 
the stonework. In some cases eye bolts were set in concrete spandrel 
beams spaced about 4 ft. 6 in. on centers, through which 1 inch base 
were placed. Anchors were carried from each stone and fastened 
around these bars. In other cases anchor bolts were anchored in the 
concrete and carried up between the joints of the limestone. Each 
anchor bolt was provided on the top with a plate and nut which hold 
the stones in place. 

INTERIOR FINISH 


The finish of concrete which is exposed to view in the interior 
of buildings is open to as great a variety of treatments as is the 
case in exterior work. Much has been done in the way of unusual 
and interesting finished surfaces; but in the great majority of con- 
crete buildings the architectural treatment of exposed interior con- 
crete has not received the attention that has been given to exterior 
work. ‘This is true because, except for the simplest possible handling 
of the material, other mediums in common use are as available and 
as readily applied as in any other type of building construction. 

For purposes of discussion, the interior finish of concrete buildings 
may be divided into the following classes: 


(1) Simple finish. 
(2) Special finish. 
(3) Concrete painted. 
(4) Plaster finish. 


3 Stone and Webster, Builders, Sanford E. Thompson, Consulting Engineer. 


756 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


Simple Finish.—Buildings in which the concrete structure is 
allowed to show on the inside are generally of the utilitarian type, in 
which good practice demands no more than a relatively clean, smooth 
surface. There are usually two kinds of defects in the concrete as 
it comes from the forms, which require attention to produce a satis- 
factory finish. These defects are the “ fins,” left projecting from 
the face where the mortar has oozed through the joints in the form-_ 
work, and the “ voids,” caused by a lack of homogeneity in the mix- 
ture and due usually to improper consistency of the mix or to insul- 
ficient spading as the concrete is deposited. 

In ordinary practice, it is usually sufficient to smooth the projec- 
tions by hammering and to fill the voids by rubbing in a rich cement 
mortar, using a wood or cork float or carborundum stone according 
to the degree of fineness desired. For a really smooth surface, the 
methods described in the discussion of exterior treatment may be 
used. | 

If the work is to be painted, it is essential, in filling the voids and 
smoothing, that no appreciable thickness of skim coat of mortar 
be left over the face of the concrete work, as such a coat is likely 
to peel under paint and flake off. 

The choice of materials for formwork and the quality of work- 
manship used in placing it have an important bearing on the elimina- 
tion of labor in finishing. 

For all work except that which is intended to be left in a rough 
state, the wood for forms, against which the concrete is to be poured, 
should be planed to bring the boards to an even thickness and to 
eliminate the marks of the grain to prevent ridges, which are sure to 
develop when boards of even slightly differing thicknesses are used. 
The fitting together of the forms should be done with care, to elim- 
imate as far as possible the cracks through which the mortar can 
flow. Knot-holes should be covered with tin or other suitable thin 
material, and where the finished surface is not to receive plaster the 
forms should be oiled. | 

Steel formwork also requires care in handling and placing, if a 
workmanlike finish is to be obtained. : 

Figure 266, p. 757, shows a ceiling surface resulting from the 
use of ordinary wood formwork and a column and column head cast 
in steel forms. 

Improvement in the manufacture and setting of steel formwork 
has advanced to the stage where concrete as it is left after the forms 


INTERIOR. FINISH 757 


have been removed may approach the appearance of a plastered 
surface; but this is accomplished only with unusual care and special 
forms, and consequently with added expense. 

Special Finish—For more elaborate work than is commonly 
expected in the average concrete building, many devices may be 
used to produce attractive and unusual results. The forms may be 
left rough, purposely showing the grain of the wood as prominently 
as possible, to be treated later with paint to imitate a wooden surface. 





Fic. 266.—Factory Interior, Salada Tea Building, Boston, Mass. (See p. 756.) 


Plaster casts may be introduced into the forms, particularly into 
the sides and bottoms of beams, to give a desired pattern or design. 
In this case, relatively fine aggregate must be used for the concrete 
which is to come next to the cast. 

The wooden formwork may be chosen for kind and quality, and 
may be left in place, solidly anchored into the concrete, when the 
mixture is poured. 

The possibilities of interesting architectural effects, to be pro- 
duced in ways similar to those mentioned above, are many and will 
suggest themselves to the resourceful designer, not only as to methods 
to be used but how and when to use them. 


758 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


Painted Concrete.—Painting of interior concrete surfaces may 
be done in the same manner and with the same materials as would 
be used on ordinary plaster surfaces, except that it is advisable to 
precede the painting with a non-burning sizing coat which is especi- 
ally prepared for use on concrete surfaces and which can be obtained 
from most paint manufacturers. The object of such a sizing coat 
is to prevent chemical action between the concrete and the paint, 
which, if not provided’ against, causes unsightly spotting of the 
painted surface. 

It must not be assumed that the use of paint will hide the form 
marks or cover up other irregularities. In fact, such treatment, 
particularly if a glossy finish coat is used, will often make the faults 
in the surface appear more marked than if left unfinished. Painting 
does, however, give a surface which is more easily cleaned than if 
the concrete were left bare, and permits color treatments superior 
in appearance to the untreated concrete. | 

Either lead and oil paint or lithopone paint is satisfactory for 
use on concrete surfaces. Ordinary cold-water paint is also used 
extensively in industrial work for ceilings, and for walls above the 
height at which the paint is likely to be rubbed off, that is, 6 or 7 ft. 
above the floor. 3 

In all painted work, the concrete should first be cleaned of any 
oil which may remain on the surface, due to the oiling of forms. 

Plaster Finish.—The use of plastering as a finished surface is as 
much an accepted form in concrete construction as it is in any other 
type of building construction, but the problem of obtaining a satis- 
factory bond between plaster and concrete continues to be an impor- 
tant one. Concrete which is to be plastered should come from the 
forms as porous as sound construction will permit, and the use of oil 
or grease on the formwork should be avoided. The plastering of 
concrete ceilings requires unusual care, in order to avoid bond failure 
with the consequent dropping of sections of plaster finish. 

It is probable that the only method of securing positive results 
in this direction is that of furring and wire lathing the surface as the 
foundation for the plaster work. This is an expensive treatment, 
and common practice does not require its adoption except in the 
most important cases. 

There are on the market preparations in the form of specially 
prepared plaster which may be applied to the concrete surface as a 
foundation for the finished plastering, and which provide a reason- 


INTERIOR FINISH 759 


ably satisfactory bond. An accepted practice for ceiling work is 
to prepare for the first or bonding coat by roughening the smooth 
portions of the concrete surface, by picking or hammering, and to 
make sure that if oil has been used on the forms it is thoroughly 
cleaned off. The amount of roughening varies widely, ranging from 
a surface hammered practically all over to hammered or picked spots 
of an area the size of the palm of the hand and spaced 3 to 6 in. 








Fic. 267.—Main Office in Salada Tea Building, Boston, Mass. (See p. 760.) 


apart. It is obvious that the more thoroughly the roughening is 
done, the less is the likelihood of bond failure. 

It is of the greatest importance in connection with the plastering 
of ceiling surfaces that the plaster be applied in as thin coats as 
possible, and in no case should more than two coats be used. The 
plastering of vertical surfaces does not present as serious difficulties 
as are met in ceiling work, average concrete being a satisfactory base, 
particularly if the special plaster bonding coat mentioned above is 
used. Rough formwork, brushing the concrete while it is still 
green, or other similar methods may be used, where practical, to 


760 ARCHITECTURAL TREATMENT OF CONCRETE BUILDINGS 


produce vertical surfaces suitable for plastering without the necessity 
of special treatment. 

Ornamental plaster work may be carried out on concrete members 
by the same method as commonly ased in connection with other forms 
of construction. It is important, however, that no molded work 
be done on horizontal surfaces without furring or other suitable 
anchorage. at 

An interesting example of the interior treatment of a reinforced 
concrete building is shown in Fig. 267, p. 759, and this may be con- 
trasted with a view of the interior of another story of the same 
building, Fig. 266, p. 757, where the construction is frankly exposed 









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to view with no further treatment than painting. In this particular 
building, the construction is of the flat slab type with the usual 
flaring column heads. ey 

The room shown in Fig. 267, p. 759, is in the first story of 
the building and is used as an office. The ceiling was furred down 
about one foot, concealing the sprinkler piping, and at the same time 
bringing the ceiling level down to a point on the flaring head of 
the column where the diameter of the structural head is such that 
a correctly proportioned column capital of cast plaster could be 
formed around it. (See Fig. 268, p. 760.) The columns themselves 
are encased in scagliola or imitation marble, with molded bases of 
the same material and plinths of dark marble. The molded ceiling, 


INTERIOR FINISH 761 


carried on channel iron furring, is hung from the bottom of the 
structural slab of the second floor, the anchorage being provided 
by hangers imbedded in concrete or sockets for holding threaded rods, 
installed in place on the formwork and incorporated in the slab when 
it was poured. For details of suspended ceiling, see p. 616. 3 

A much simpler treatment of a circular column head is shown in 





Fic. 269.—Office Building Interior, Columns, Walls and Ceilings Plastered. 
(See p. 761.) 


Fig. 269, p. 761. The plaster molding at the top cf the capital in 
this case is run without furring, as it approximates the controur of 
the simple molding of the steel form. 

It is obvious that, in beam and girder construction, the use of 
cornices, capitals, moldings, etc., is relatively simple, provided that 
such furring is used as is ordinarily employed under liké conditions in 
other types of fireproof construction, 


CHAPTER XVIII 


CONCRETE IN CONSTRUCTION OF THEATERS AND AUDI- 
TORIUMS 


Reinforced concrete is well adapted for theater construction. 
It has been used for this purpose in a large number of cases, in suc- 
cessful competition with steel construction. In localities distant 
from large steel fabricating plants, it 1s particularly economical, 
since the materials for reinforced concrete theater construction are 
more easily obtainable than structural steel members. It is also 
useful for baleony cantilevers in combination with steel trusses, 
whereby the difficult connections between the steel cantilevers and | 
trusses are avoided. | | 

The theater, for the sake of discussion, may be divided into the 
following parts: orchestra floor, balconies, proscenium, and roof. A 
general idea of a design of concrete theater may be obtained from | 
Fig. 270, showing a section through Grauman’s Theater Building in 
Los Angeles and Fig. 275 showing cross section through Winston- 
Salem auditorium. , | | 3 


ORCHESTRA FLOOR 


The space under the orchestra floor is often entirely unexcavated 
or excavated only in part. When the ground is solid enough, the 
concrete floor for the unexcavated portion may rest directly upon it 
and be treated in same fashion as the ground floor in ordinary build- 
ings. To prevent cracking due to contraction and shrinkage, some 
reinforcement, consisting of small bars or mesh, is advisable. The 
orchestra floor in the Winston-Salem auditorium, shown in Fig. 
275, p. 781, rests directly on the ground. 

If the space under the floor is excavated, or if it is not desirable 
to rest the slab on the ground, the construction of the orchestra floor 
consists of one-story columns supporting a reinforced concrete floor 
system. It differs from ordinary building construction in that the 

762 


BALCONY DESIGN 763 


slab is level only near the stage, and then slopes up. In many cases 
the slope is gradual at first and then more pronounced. The longi- 
tudinal cross section is then a curve with a horizontal tangent at the 
origin. When the rows of seats are parallel to.the stage, all cross 
sections are alike. The only difference between such construction 
and ordinary building construction is that the columns in successive 
rows will increase in height and the beams or girders longitudinal 
with the building will be inclined, while the members across the 
building will be level. 

Ordinarily, however, the seats in each row are placed on a curve 
which is either a segment of a circle or, more often, a three-centered 
curve. Since all seats in a row must be on the same level, the orches- 
tra is not a plane but a warped surface. This complicates the floor 
construction. The best solution is to place inclined girders longi- 
tudinally with the building, and then arrange the beams tangentially 
to the curve of the seats. All beams in a row will then be on the same 
level. 

Unless fixed by the requirements of the space below the orchestra 
floor, the spacing of the columns carrying the orchestra floor will 
depend upon economic considerations, and particularly upon the cost 
of foundation and length of columns. If the cost of foundation for 
each column is small and the columns short, they may be spaced 
more closely. For difficult foundation work, larger spacing of 
columns will be found economical. Usually, a spacing of 18 to 20 ft. 
will be found satisfactory. 

Although beam and girder construction is generally used for 
orchestra floors, flat slab construction in many cases is more economi- 
cal. Either type of floor is designed according to the principles 
discussed on pp. 303 and 575. The cross section through Grauman’s 
theater, Fig. 270, p. 764, shows a design of supported orchestra floor. 
In this case the construction is of the beam and girder type with 
girders running longitudinally with the building. 

The steps for the seats may either be made integral with the 
floor slab or be built separately of cinder concrete on top of the 
structural slab. 


BALCONY DESIGN 


The most difficult problem in theater construction is to design 
the balconies without using intermediate columns in the orchestra. 
This is usually effected by the introduction of one or more fulcrum 


CONSTRUCTION OF THEATERS AND AUDITORIUMS 


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girders running across the theater and serving as supports for the 
inclined cross beams carrying the balcony slab. As it is impossible 
to place the fulcrum girders at the edge of the balcony, the inclined 
cross beams must be cantilevered out. often to a considerable length. 


FULCRUM GIRDERS 


Arrangement of Fulcrum Girders—The arrangement and the 
number of fulerum girders depend upon the width of the theater. 

The simplest arrangement is shown in Fig. 271 a, p. 766. One 
fulcrum girder is placed as near the edge of the balcony as possible. 
It is supported on columns concealed in the wall, or, if this 1s possible 
without obstructing the view, a few feet from the wall. The latter 
arrangement is practicable, for example, when columns can be con- 
cealed in boxes or placed at the edge of outside aisles. Full-span 
fulerum girders are particularly adaptable for small spans. They 
have been used, however, for spans as large as 126 ft. 

The balcony seats are usually placed on a curve. With a fulcrum 
girder placed across the theater, as in the first arrangement, the 
cantilever arms at the sides may be too long. They may be reduced 
by using side diverging girders, one on each side; one end of each 
resting on a column placed in the side wall and the other end on the 
main fulerum girder. (See Fig. 274, p. 780.) 

Where the width of the theater is large, it may be economical to 
use the more complicated arrangement of fulcrum girders shown in 
Fig. 271 b, p. 766. Here two side girders are supported by four 
columns, and a middle girder spans between the side girders. This 
has the advantage of requiring spans not larger than 50 or 60 per 
cent of the width of the theater. It is economical to make the span 
of the side girders smaller than that of the middle girder. The aggre- 
vate leneth of the three girders in this arrangement is greater than 
the length of the one girder in the arrangement previously described, 
but their size is much smaller. The required depth for the middle 
eirder is much smaller than would be required for a girder spanning 
the whole width of the theater, and it can be placed, therefore, 
nearer the edge of the balcony and thus reduce materially the canti- 
lever arm of the balcony beams. 

Position of the Fulcrum Girder.—Several considerations affect 
the position of the fulcrum girder in relation to the edge of the balcony. 
The first is the depth required for the girder. The available depth 


766 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


‘3 the distance between the top of the balcony floor and the line fixed 
by the minimum headroom for the seats below. At the edge of the 
balcony, this available depth is made as small as possible, and usually 
equals only a few inches. Since the balcony seats are stepped up, 
the available depth increases rapidly with the distance from the 
balcony edge. As far as the economical design of the girder alone 
is concerned, it would be placed where the available depth is equal 
to the economical depth for the span of the girder. 

The second consideration is the design of the balcony cantilevers 
carried by the fulcrum girders. The length of the cantilever arms 







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obviously depends upon the distance between the fulerum girder 
and the balcony edge. As explained in Vol. II, the cost of the canti- 
lever is a minimum if the length of the cantilever arm is about equal 
to one-half of the anchor arm. It will be found, in most cases, that 
the two considerations are conflicting, and the problem is solved by 
considering the economy of the construction as a whole. Usually, 
it will be found economical, for the construction as a whole, to place 
the girder at a distance from the edge where the available depth 
equals the minimum depth for which a girder of the required span 
can be designed. The sacrifice in the economy of design of the girder 
will be more than offset by the economy in cantilever beams. 


FULCRUM GIRDERS 767 


Shape of Fulcrum Girders.—The fulcrum girders may be either 
T-beams, trusses, or arches. The design of each type is discussed 
below under the proper heading. 

Fulcrum Girders as T-beams.—The slab in the balconies is thin, 
and cannot serve as a flange for the girder. If a rectangular beam 
has not sufficient compression area, a special compression T-flange 
must be provided and the fulcrum girder built as shown in Hign27 7, 
p. 783. If possible, the flange should be large enough to make 
compression steel unnecessary. For allowable width of the flange 
see p. 218. The flange should be as symmetrical as possible 
about the center line of the girder. In most cases, the width of 
flange on the balcony side is limited by the shape of the balcony, but 
on the opposite side any width is available. For this reason some 
designers use a flange appreciably wider on one side than on the other. 
The authors do not approve of this, because in unsymmetrical beams 
the neutral axis is inclined and stresses in steel are not uniform. Spe- 
cial formulas would have to beused for unsymmetrical beams. In the 
design on p. 783, the flange was made symmetrical. The total width 
was fixed by the available width on the balcony side. This width 
was not sufficient, and therefore compression steel was used. 

The web of the T-beam is fixed by the shear and also by the 
required width for the placing of the large number of bars. Some- 
times it is economical to make the web wider near the supports, 
where the shear is a maximum. Also, the dead load of the beam 
may be reduced by paneling the web in the central portion of the 
beam. 

In designing the girder, especially where the flanges are thin in 
comparison with the depth of the girders, it is advisable to use the 
formulas in which the compression stresses in the concrete below the 
flange are taken into account. A simple exact method is given on 
p. 224. 

The fulcrum girders are almost always of single spans. Since 
the columns supporting them have a much smaller moment of inertia, 
it is not possible to restrain the ends of the girders to any extent. 
Therefore, they should be considered as simply supported, but top 
reinforcement at the support should be provided in sufficient amount 
to resist any negative bending that may be developed. 

Special attention should be paid to the design of reinforcement. 
To accommodate the large number of bars, an unusually large number 

of layers are required. To develop full stresses in a large number of 


768 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


bars placed in several layers with a comparatively small amount of 
concrete between the layers, the rules given below must be strictly 
followed: 

First-—The bond stresses must be kept within working limits, 
while at the same time, to reduce the number of bars, their diameters 
should be made as large as permitted by bond stresses. 

Second.—The bars in each row must be kept in place by positive 
means. The rows of bars should be separated by short bars of 
proper diameter placed across the main reinforcement between the 
layers, and spaced about 2 or 4 ft. on centers. The diameter of these 
cross bars must be equal to the diameter of the largest bar, but must | 
not be less than one inch. 

Third.—From 50 to 60 per cent of the number of bars should be 
bent up and carried to the top of the beam. ‘The points of bending 
should be determined from the bending moment and well distributed. 
“Some of the bent bars should be extended along the top to the end 
of the beam. All of the bars should be provided with hooks at their 
ends. The bending of the bars serves two purposes: (1), the bent 
portions act as diagonal tension reinforcement; and (2), by anchoring 
the trussed bars in the compression zone the development of their 
strength is insured. Since the bent bars are distributed through the 
beam, a large mass of concrete assists in developing the tensile stresses 
in the bars. The number of the straight bars depending upon bond 
alone is materially reduced. 

All straight bars should be provided with hooks at the ends of the 
beam. 

Fourth.—Stirrups should be used throughout the whole length of 
the beam, and kept in place by bars placed in the top of the 
beam. 3 

Fifth.—The T-flange of the girder should be provided with cross 
bars having a total cross-sectional area of at least 0.2 per cent of the 
section of the flange. These should be supported and kept in place 
by longitudinal bars placed in the flanges. 

Sizth—Under no circumstances should the longitudinal rein- 
forcement be reduced simply by stopping bars at points of reduced 
bending moment. Designs with a number of straight bars shorter 
than the span have been used, but cannot be condemned too strongly. 
It should be evident that stresses in the short bars cannot be induced 
without slipping of the bars (see p. 289). This would be doubtful 
practice even in small beams with one or two layers of bars. In 


FULCRUM GIRDERS 769 


large beams, no amount of saving in steel that may result from this 
arrangement is sufficient to warrant the bad practice. 

Seventh.—If there is any possibility of negative moment, top 
reinforcement, properly anchored at ends, should be used even if no 
advantage is taken of the restraint in computing the positive moment. 

Fighth—Columns supporting the girders should be designed to 
take care of any bending moment. 

Fulcrum Girders as Trusses.—For large spans, the amount of 
concrete may be reduced by using concrete trusses instead of T-beams. 

fectangular Truss.—The trusses may be obtained simply by 
providing rectangular openings throughout the girder. The truss 
would then consist of chords and vertical members, except at the 
ends, where solid web may be used for the distance required to 
provide for shear. ‘This type of construction is statically indetermi- 
nate. ‘The chordsand the vertical members are subjected not only to 
direct force but also to bending. The bending stresses in these mem- 
bers depend upon the external shear and are larger near the end than 
in the middle portion of the truss. Method of design of trusses 
is given in Volume III. | 

Triangular Truss—To avoid the large moments in the members 
due to shear, concrete trusses are often designed with diagonals. 
_ The truss consists of triangles, and the direct stresses may be com- 
puted in the same manner as in steel trusses. 

In addition to direct stresses due to'truss action, all bending 
stresses due to the dead load of the members, and to any load placed’ 
between the panel points, should be computed. In computing 
bending moments, the members may be considered as continuous 
or fixed beams supported at the panel points. The stresses due to 
the bending moments should be combined with direct stresses as 
explained on p. 187. 

Because of the rigidity of the joints, secondary bending stresses 
will develop. These, however, are no larger than in a steel truss 
under similar conditions. In the shallow steel trusses commonly 
used in theater construction, the rigidity of joints, due to heavy 
gusset plates, is about the same as in concrete construction. In 
both cases, proper allowance ‘should be made for secondary stresses. 
Reinforcement in the truss members should be placed so that it is 
able to resist tension due to bending moments to which the members 
may be subjected. 

In designing the top and bottom chords, it must be remembered 


770 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


that the computed stresses will not be uniformly distributed over the 
whole sectional area of the member. T he depth of the chords is 
large in comparison with the theoretical depth of the truss. After 
deflection, the top chord will shorten more at the top edge than at 
the bottom edge, although for chords of small depth this difference 
in deformation is small. In theater construction, owing to the large 
depth of the chords, the difference in shortening of the two edges is 
appreciable. The resulting variation in stresses is proportional to 
the variation in the deformations. The member must then be 
designed for an average unit stress smaller than the maximum allow- 
able stress. ‘The same principle applies to the stresses in the bottom 
chord. 

Allowable Compression Stresses in Trusses.—Since the compres- 
sion members in trusses are subjected to direct compression (similar 
to column stresses), the same unit stresses allowed for columns 
should be used, and not the stresses allowed for extreme fibers in 
flexure. However, when bending moments are considered in addi- 
tion to direct stresses, the formulas on p. 170 should be used and the 
allowable combined stress may be accepted as explained on p. 463. 

Design of Tension Members in Trusses.—In tension members, 
all the tensile stresses are resisted by the reinforcement, and the 
area of concrete is made only sufficient to properly cover the bars. 
The same spacing of bars should be used as recommended on p. 273. 7 
While a small number of large bars is easier to handle and makes a 
compact tension member, its use is recommended only in cases where 
the tension member extends the full length of the truss, where any 
splicing is accomplished by mechanical means, such as turnbuckles, 
and finally where the stresses are transferred to the concrete at the 
ends by nuts and anchor plates. When the tensile stresses in 
bars are developed by bond, a large number of small bars is advocated. 
In addition to direct tension, bending, due to dead load, and second- 
ary bending stresses must be considered. In some cases, bending 
moments are so large that the concrete may assume the duties of 
resisting compression stresses produced by bending. Formulas given 
on p. 191 should be used. 

Usually, however, the bending stresses are not sufficient to change 
the tension, due to direct stresses, to compression. 

The proper design of the details and connections between members 
is of great importance. Bars should be spliced only at panel points. 
To avoid crowding of bars, as few splices as possible should be used. 


FULCRUM GIRDERS 771 


The bars should extend over several panels and should be arranged 
so that only a few bars are spliced at any one panel point. At the 
joints, the bars of one member must extend far enough into: the 
adjoining member (or the intersection of adjoining members) to 
develop the full strength of the bar, taking into consideration, of 
course, the character of the stresses to which the bar is subjected. 
A bar effective in one panel of the bottom chord, for instance, should 
extend beyond the panel point into the adjoining panel for a length 
equal to 40 diameters of the bar for deformed bars, and to 50 diam- 
eters for plain bars. The solid concrete at the joint, which cor- 
responds to the gusset plates in steel construction, must be of suffi- 
cient size to enable the concrete there to transmit the stresses from 
any one member to the other members in the joint. 

Each joint, in addition to the reinforcement of the members 
extending into it, should be provided with special horizontal and 
vertical reinforcement in the shape of hoops or stirrups. These 
serve to take care of any secondary stresses and preserve the integ- 
rity of the joint. 

End Panel Point.—At the end panel points, proper provision 
should be made for diagonal tension due to the external shear. The 
structure may fail by forming a diagonal crack between the com- 
pression and tension members. To prevent this, the vertical area 
of concrete at the junction of the two members must be sufficient to 
keep the unit shearing stresses, computed as in ordinary beam design, 
from exceeding the maximum allowable working values. In addi- 
tion, the authors recommend that the minimum horizontal cross 
section of the joint should be equal to the reaction divided by 100 lb. 
_In this distance, the cross section of vertical stirrups should be equal 
to the reaction divided by 16000 lb. Stirrups must be looped 
around the tensile bars at the bottom and around the compression 
bars at the top. 

All tensile stresses are resisted by the reinforcement. The area 
of concrete is made only large enough to cover the bars. The tensile 
reinforcement in the bottom chord, if it consists of large bars, should 
be provided with a screw and nut at the end and an anchor plate 
of sufficient size to transfer all the stresses from the bar in bearing 
on the concrete. Reinforcement consisting of small bars should be 
provided with hooks at the ends. If possible, the tensile bars should 
hook up into the compression member. If not, they should hook 
around the reinforcement of the compression member. 


772 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


Relative Economy of T-Beam and Truss.—The relative economy 
of the T-beam and the truss for fulcrum girders depends upon the 
cost of formwork. The truss has much less concrete and steel, but 
the formwork is complicated and the placing of concrete and steel 
more difficult. In many instances, the truss without diagonals, | 
(that is, with rectangular openings) even if it is necessary to use 
solid web at the ends, may prove more economical than the lighter 
truss with diagonals because of cheaper formwork. In cases where a 
decided saving cannot be demonstrated, T-beam design should be 
used in preference to the truss as it 1s easier to build. 

Fulcrum Girders as Arches.—Arch construction is well adapted 
for fulerum girders for large spans. As it is not feasible to make the 
support unyielding enough to take the horizontal thrust, the arch 
must be provided with tie-bars of sufficient area to resist the 
thrust. If properly designed, the cost of the arch is smaller than the 
cost of the truss under similar conditions, mainly on account of the 
simplified formwork. 

For designing purposes, the arch should be considered as hinged 
at the supports, which makes the horizontal thrust the only statically 
indeterminate quantity. 

The cantilever beams may be placed between the top of the arch 
and the bottom level of the tie-bars. Tension bars of sufficient 
area to suspend the cantilever from the arch must then be used. 

The method of design is discussed in Volume III. 

Special Concrete for Fulcrum Girder.—To save dead load, it may 
be economical to use richer mix of concrete for fulerum girders. A 
mix of 1:13:23 or even 1: 1:2 may prove economical, especially 
when, with the leaner mix, compression reinforcement would be 
required. 


BALCONY CANTILEVERS 


The balcony floor slab is carried by beams, which in turn rest on 
the fulerum girders. As already explained, the fulcrum girders 
must be placed some distance back from the edge of the balcony, 
and therefore the balcony beams must be cantilevered out beyond 
the girder. 

Reinforced concrete is well adapted for the balcony cantilevers 
and is being used for this purpose, not only in reinforced concrete 
theaters but also in connection with structural steel fulerum girders. 


BALCONY CANTILEVERS 173 


As a general proposition, the cantilever beam carrying the 
balcony consists at one end of the cantilever arm, resting on the 
fulcrum girder, and at the other end of span the anchor arm, anchored 
to the support to prevent uplift. In modern theaters, the length of 
the balcony may be considerable. With only one fulcrum girder, 
the length of the cantilever arm may exceed 30 ft. 

If possible, the cantilever arm should not be longer than the 
anchor span. In no case should the uplift caused by the load on the 
cantilever arm (multiplied by proper factor of safety) with the anchor 
span unloaded, exceed the available downward reaction at the point 
of anchorage. 

Moments and External Shear in Cantilever.—The beams must be 

designed for three assumptions as to the disposition of the live load, 
each of which produces maximum stresses in some part of the con- 
struction. The assumptions are as follows: 

1. The cantilever arm is loaded, the anchor span not loaded. 

2. The cantilever arm is not loaded, the anchor span loaded. 

3. Both arms are loaded. 

The first condition, where only the cantilever arm is loaded, 
produces maximum stresses in the cantilever arm and also maximum 
negative bending moments in the anchor span. The uplift is also a 
maximum for this condition. 

The second condition, where the anchor arm ant is loaded, 
produces maximum positive moments in the anchor span. 

The third condition, where both arms are loaded, produces maxi- 
mum shears in the anchor span and also maximum reactions on the 
fulcrum girder. 

The moments and shears for the live load should be combined 
with those for the dead load. For some conditions of loading, the 
stresses produced by the live load are of opposite sign to those pro- 
duced by the dead load, and if the two are added, the dead load 
stresses may balance partly or fully the live load stresses. Under 
such conditions the combination of live load stresses and dead load 
stresses by a simple addition, is not permissible, but the live load 
stresses must be multiplied by the required factor of safety. When 
the live load is doubled, the stresses produced by it are doubled also. 
The dead load stresses, however, remain the same, so that the reduc- 
tion in the live load stresses due to the dead load stresses is propor- 
tionally only half as large for double the live load. 

The following example explains the matter more fully. Assume 


774. CONSTRUCTION OF THEATERS AND AUDITORIUMS 


that the anchor span is 20 ft. long, cantilever arm 12 ft., uniformly 
distributed dead load 1 000 lb. per lin. ft., uniformly distributed live 
load 1 400 lb. per lin. ft. 

Dead Load Reactions: 


Due to load on cantilever arm: 


At fulcrum girder, lett. 5. eee 12 000 Ib. downward 

At fulcrum girder, right.......... 3 600 lb. downward 

At. end SUuDpOTt ses ace oe .. 8600 lb. upward 
Due to load on anchor span: 

At fulcrum girder, right.........-. 10 000 Ib. downward 

At'end SupDORisie- = saa -~. 10000 Ib. downward 


"The sum of the two reactions gives: 
Reaction due to dead load: 
At fulerum girder, left.........7¢5: 12 000 lb. downward 
At fulcrum girder, right, 
10 000 + 3 600 = 13 600 lb. downward 
At end support.. .10 000 — 3600 = 6 400 lb. downward 
No uplift due to dead load. 


Live Load Reactions, Cantilever Arm Only Loaded: 


At fulerum girder, right. .......... 14000 lb. downward 
At fulcrum girder, left...2 72-1. eee 5 040 lb. downward 
Atend'support: 25.20. 73 5 040 lb. upward 


Uplift due to live load, 5 040 lb. 


Combining the uplift due to the live load, equal to 5 040 Ib., with 
the downward reaction due to the dead load, equal to 6 400 lb., it 
will be found that no uplift will occur for this condition of loading, 
and therefore no anchorage would seem to be necessary. But in 
such a case, practically no factor of safety against uplift would exist. 

If, now, the live load on the cantilever is doubled, the uplift due 
to the live load will also be doubled, or 2 X 5 040 = 10 080 lb. The 
downward reaction due to the dead load will remain the same. The 
net uplift, therefore, for double live load on the cantilever, will 
amount to 10 080 — 6 400 = 3 680 lb. mtr | 

To get a factor of safety of 2 against uplift, it is necessary to 
anchor the structure strongly enough to resist an upward pull of 
3 680 Ib., even if for the combination of single live and dead load a 
downward reaction is obtained. 


BALCONY CANTILEVERS 115 


The same applies to the stresses due to negative bending moment 
in the anchor span. To provide a proper factor of safety, it may be 
necessary to provide negative moment reinforcement in places where 
the combination of one live load and dead load give positive moment. 
This difference between ordinary beam design and cantilever design 
must be constantly borne in mind. 

To combine dead and live load moments when they are of oppo- 
site sign, the procedure is as follows: The moments due to live load 
are multiplied by the desired factor of safety. ‘The moments due to 
dead load are subtracted. The difference is then divided by the 
same factor of safety and the dimensions of the member determined 
in the usual fashion. 

In many parts, a reversal of stress may take place; that is, for 
some loading the stress will be tension, and for other loading, com- 
pression. The frequency of the reversal is not sufficient to require 
any reduction in unit stresses. | 

Depth of the Cantilever Beam.—In most cases, for designing 
purposes, the cantilever beam is considered as a rectangular beam. 

The required depth of the cantilever beam is governed either by 
shear or by the negative bending moment, both of which are a maxi- 
mum at the fulerum girder. 

In the cantilever arm, because of the slope of the balcony, the 
beam decreases in depth, and this decrease may be larger than the 
decrease in bending moments and shears. Therefore, it is not suffi- 
cient to determine the depth and the amount of steel at the support; 
it is also necessary to investigate the stresses at intermediate points 
and design the sections to correspond. 

In the anchor span, the shears and bending moments are largest 
at the fulcrum girder. The depth of this girder, therefore, is made 
the same as for the cantilever arm at this girder, and the beam is then 
tapered toward the other support. The taper may be uniform, or a 
fairly steep hunch may be used at the support followed by a gentle 
taper toward the end of the beam. 

The depth of the cantilever beam must be the same on both sides 
of the fulerum girder, so that the compression stresses on one side 
may be resisted by compression stresses on the other side of the 
girder. If the depth were different, torsion would be developed in 
the girder. In some designs, the moment produced by the cantilever 
arm is resisted on the other side of the girder by a tie beam on the 
top and a strut at the bottom. It is obvious that in such designs 


776 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


no provision is made for the shear produced by the bending moment, 
and as a result heavy secondary stresses are developed in both 
members. 

Reinforcement for Cantilever Beams.—The reinforcement deter- 
mined for the negative moment in the cantilever beam at the fulcrum 
girder must be extended on both sides. In the cantilever arm, at 
least one-third of the bars should extend on the top for the full 
length of the arm; the rest may be bent down to resist diagonal 
tension. In determining the points where it is permissible to bend 
the reinforcement, it is necessary to remember that, while the bend- 
ing moments decrease with the distance from the support, the depth 
of the cantilever decreases. For steep balconies, the decrease in 
depth may be almost as rapid as the decrease in bending moments, 
-n which case no reduction in negative steel is permissible and all 
bars must extend on the top for the full length of the cantilever arm. 

In the anchor span, some of the top steel from the cantilever arm 
should be carried on the top, for the full length of the anchor span, 
because, for the most unfavorable position of the load, negative 
moment will be found throughout the span. This happens when the 
stresses due to live load on the cantilever arm only are multiplied by 
the factor of safety and are combined with the dead load stresses. 
The rest of the bars may be bent down, when permitted by the 
moments, and extended at the bottom for use as positive moment 
reinforcement. Sometimes, the cantilever is of such shape that it 1s 
difficult to extend the bars along the top from end to end without 
difficult bending. In such cases, bars may be spliced, provided 
ereat care is used to break joints. Some steel should be used at the 
bottom of the cantilever span, even if not required for compression. 

Positive bending moment reinforcement in the anchor span 
should be treated as in ordinary beams. 

Cantilevers in Steel Construction.—Concrete balcony cantilevers 
may be used with economy in connection with structural steel ful- 
crum girders. In all cases, the girders are deep enough to allow 
sufficient depth for an economical concrete member. The economy 
effected by the use of concrete instead of steel consists not only in the 
saving in the cost of the member itself, but also in the simplifying of 
the connections between the girder and the cantilever. 

The design of a cantilever supported by a steel girder does not 
differ to any extent from that of a cantilever supported by a rein- 
forced concrete girder. The only interesting part is the method of 


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7783 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


supporting the cantilever on the steel truss. In Fig. 272 is given acan- 
tilever used in the theater auditorium of the State-Lake Building in 
Chicago.! The part of the supporting girder is also shown, as well 
as the detail of the seat for the cantilever. The concrete poured 
between the gusset plates and resting on the bottom chord, serves 
as a seat for the cantilevers. 

Cantilever Trusses for Balconies.—Ordinarily, cantilevers are 
rectangular beams. For large spans, trusses have been used, as shown 
in Fig. 278, p. ¢#9.- 


BALCONY FLOOR CONSTRUCTION 


The top of the balcony is stepped. The most common construc- 
tion consists of vertical ribs, spanning from cantilever to cantilever 
and forming the risers of the steps, and a thin slab spanned between 
the ribs. The thickness of the slab is usually 3 in. The ribs are 
about 5 in. thick, and their depth is governed by the required height 
of riser for the step. The ribs must be designed as beams, each 
carrying half of the load on the two adjoining slabs. The slabs 
should be reinforced with bars extending into the ribs. Although 
negative moment will be developed in the slab, it is not practicable 
to bend up the slab steel. Wire mesh is sometimes used as slab 
reinforcement. The top of the cantilever extends to the top of the 
floor and is also stepped. 

Another method sometimes followed is to use an inclined slab 
between the cantilever beams (or inclined slab and joist construction) 
and then build the steps of lean cinder concrete built on the top. 
In this way the formwork is simplified, but the construction may 
prove heavier and more expensive than in the previous case. 


ROOF CONSTRUCTION 


The roof construction may consist either of simple girders or of 
trusses. Arches with ties may also be used. The long span roof 
construction is discussed in Chapter XIII. 


THEATER AT WINSTON-SALEM, N. C. 


Examples of Theater Construction—A good illustration of the 
adaptability of reinforced concrete in theater construction. is shown 


1 For description, see Engineering News-Record, Vol. 83, July 24, 1919, p. 178. 
2 For full description, see Engineering News-Record, July 5, 1923. 


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THEATER AT WINSTON-SALEM 


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784 CONSTRUCTION OF THEATERS AND AUDITORIUMS 


in Fig. 274 to 277, taken from the design of the theater in Winston- 
Salem, N. C., designed by W. C. Northup, Architect. 

The structural plans were prepared by The Southern Engineering 
Go., with Mr. Edward Smulski as Consulting Engineer. The plan 
of the first balcony is shown in Fig. 974. The cross section through 
the theater is shown in Fig. 275. Fig. 276 gives reinforcement in the 
first balcomy cantilevers. Fig. 977 shows the design of the fulerum 
girder. 


CHAPTER XIX 


REINFORCED CONCRETE IN DIFFERENT TYPES OF 
BUILDINGS 


In this chapter, dealing with the different types of buildings in 
which reinforced concrete is commonly used, the economy of con- 
struction is briefly discussed and the design live loads are given. 
Recommendations are made as to proper spacing of columns and 
proper type of floor construction. Floor finish is also discussed. 

The following types of buildings are considered: warehouses, 
p. 785; cold-storage warehouses, p. 792 ; manufacturing buildings, 
p. 792; buildings used for the manufacture or storage of automo- 
biles, p. 795; office buildings, p. 803; hotels, apartment houses, and 
hospitals, p. 805; schools, p. 808. 


WAREHOUSE CONSTRUCTION 


Reinforced concrete is well adapted for warehouse construction, 
and a large majority of fireproof warehouses are built of this material. 

Type of Floor Construction.—The majority of reinforced concrete 
warehouses are of flat slab construction. Beam and girder con- 
struction or joist construction is rarely as economical as the flat slab 
for this type of buildings. Joist construction should not be used for 
heavier live load than 125 Ib. per sq. {t., because of the possibility 
of heavy load concentrations for which joist construction is not 
adapted. 

An idea of the relative economy of several types of floor construc- 
tion may be obtained from the table on p. 786, taken from a paper 
by A. F. Klein.! This table gives costs of three types of floor con- 
struction for a warehouse consisting of eight stories and basement, 
with 20 by 20-ft. panels designed for a live load of 250 lb. per sq. ft. 
The following unit costs were used: 

* Analysis of Cost of Types of Fireproof Construction. By Arthur F. Klein. 
Journal Western Soc. of Engineers. Vol. XXIX, No. 7, July, 1924. 

785 


936 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


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WAREHOUSE CONSTRUCTION 787 


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1 Roof Slabs 6" yn 
Typical Floor 83 






Fic. 280.—Type 3. Flat Slab Floor for 8-Story Warehouse. (See p. 786.) 


788 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


Cost of Concrete—Cement, $2.40 per barrel net. Crushed stone 
and torpedo sand, $2.74 per cu. yd. delivered. On this basis, cost of 
1 :2:4of concrete in place is $12.50 per cu. yd., or 44 cents per cu. ft. ; 
and of 1: 1 : 2 concrete $15.00 per cu. yd., or 54 cents per cu. ft. 

Lumber for formwork figured at $47.00 per thousand feet, board 
measure. 

Reinforced Steel—Cost of steel based on $60.00 per ton, base, 
with a standard increase in price, for sizes below base. Following 
additional costs were used: 


Per ton 
Trucking: .. 2... - #1 eel th Sines = one $3.00 
Standard bending of steel.........-.+-+++++-> 6.00 
Bending stirrups. ./...5. .- =>". pe 18.00 
Shop details. .....-.4- +--+ = =e a 3.00 
Spirals... +a cen eines o> 2 90 . 00 
Placing of steel. 2... . 1417 eee 20.00 


The cost of formwork is based on the assumption that forms for 
9} floors will be provided, and therefore that they will be re-used four 
times. , 

Values given in the table are total estimated costs of an interior 
section of a building equal in area to 20 by 20-ft. panel and 8 stories 
high. The total costs are as follows: 


Cost per section Cost per 


of 8 stories. sq. ft. 
Flab slab design. ...+ +. «25+ e-0+ +++ += nie $4 054.13 $1.13 
Square panel with four hears Aer ac. 5 ce 5 320.02 1.48 
Beam and girder design. .....--+++-+++es000 5 645.24 1.54 


The cost of flat-slab design is much lower than that of beam and 
slab designs. The difference in cost between the two-beam and girder 
types is small. 

It must be noted that the above figures cannot be used for esti- 
mating the cost of the frame of an entire building as the cost of 
exterior panels will be larger, owing to heavier construction of floors, 
additional cost of spandrels, and also additional columns. In figuring 
the cost of an interior panel, the cost of one interior column only 
needs to be included, while for an exterior panel it is necessary to use 
one exterior column plus one-half of an interior column. In a corner 
panel, extra cost of corner column must be considered. 


WAREHOUSE CONSTRUCTION 789 


Spacing of Columns.—Column layout should be governed pri- 
marily by the requirements of the occupants. If no other consider- 
ation governs, the most commonly used and most economical spacing 
varies from 18 to 20 ft. The panels, preferably, should be made 
Square or nearly square. Where the shape of the building can be 
made to suit, its length and width should be multiples of the accepted 
square panel. The end panels may be made somewhat smaller than 
the interior panels, as the moment coefficients in end panels are 
larger than in interior panels. With smaller span, it is possible to 
maintain the same dimensions in exterior and interior panels. 

When the dimensions of the building are fixed by the size of the 
plot, the choice of the panel sizes is restricted. For instance, if the 
outside dimensions are 100 ft. by 120 ft., 20 by 20-ft. panels follow 
almost automatically. The 100-ft. length, divided into four parts 
instead of five, would give a panel length of 25 f{t.; divided into six 
parts, it would give a panel length of 16 ft. 8 in. 

The possible panel sizes in a 100 by 120-ft. building are 16 ft. 
8 in. by 17 ft. 1 in., 20 by 20 ft., and 25 by 24 ft. The smallest panel 
may be cheaper than the 20 by 20-ft. panel, but the building will not 
be as useful. The 25 by 24-ft. panels are too expensive. Ordinarily 
therefore, 20 by 20-ft. panels would be selected. 

If the building is not rectangular, the main column lines should be 
laid out at right angles to each other and the skewed portion composed 
of odd panels. This rule does not apply when the building lines form 
arhomb or rhomboid. In sucha case, the opposite sides are parallel, 
but the intersecting sides are not at right angles. The column lines 
then may be placed parallel to the sides of the building. 

Design Live Load.—The magnitude of the design live load 
depends upon the uses for which the warehouse is designed. No 
warehouse should be designed for a smaller live load than 125 Ib. per 
sq. ft., a live load permitted for furniture storage. This load should 
never be used if there is any possibility of a change in the ownership 
of the warehouse. For general storage,a minimum live load of 150 Ib. 
per sq. ft. is recommended. In warehouses several stories high, a 
good plan is to design the first floor for 250 to 300 lb. per sq. ft., the 
next floor or two for 200 lb. per sq. ft., and the balance for 150 Ib. per 
sq. ft. ‘The rentability of such a warehouse is much superior to that 
of one in which all floors are designed for a small live load. 

In determining the live load for a warehouse which is to be rented, 
it should be kept in mind that the adoption of too small live loads 


790. REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


may restrict the use of the warehouse considerably, and may make 
‘t less rentable, so that the small economy in construction may 
result in actual loss. The difference in cost between a warehouse 
designed for 150 lb. and one designed for 250 lb., at prices prevailing 
in 1925, is only 10 cents per sq. ft. This means an increase of about 
8 per cent in cost of the structural frame, but only 3.5 per cent of 
the cost of the complete structure. | 

In large cities, the capacity of a floor is posted. Although in many 
cases this is only of theoretical value, still there is a possibility of the 
law being enforced. A warehouse that is constantly overloaded will 
develop cracks which, even if not dangerous, are unsightly and limit 
the market value of the building. The difference in cost between a 
building with light live load and one with heavy live load is negligible 
when the larger usefulness of the latter structure is taken into con- 
sideration. 

Loading Platform.— Where possible, provision should be mide 
for railroad deliveries of goods, by the addition of a covered loading 
platform extending the full length of the track. The width of the 
loading platform may have to be made large enough to afford 
temporary storage for the goods. The loading platform should be 
designed for a minimum of 300 Ib. per sq. ft. 

Where ground is available, the side tracks are run outside of 
the building and the platform projects from the building. It is 
usually covered with a permanent canopy. In U-shaped buildings, 
the tracks may be placed in the court. | | 

In cities where the ground is very valuable, it is not possible to 
waste the space which would be required for tracks outside the build- 
ing, and the tracks are therefore run into the building. A straight 
track, running parallel to a column line, can easily be accommodated 
in one panel width. The curve required to connect the side track 
with the main track should be placed preferably in the court or out- 
side the building. If this curve is placed within the building, it is 
necessary to offset some of the columns in such a way as to give the 
required clearance for the train. Sometimes the position of the 
offset columns is maintained in all floors. Usually, however, it is 
cheaper, and gives a better job, to maintain uniform spacing of the 
columns above the second floor. The columns above the track are 
then carried by heavy girders placed below the second floor. Such 
girders will usually be of rather large dimensions. If possible, they 
should be built of reinforced concrete, even if the cost of the materials 


WAREHOUSE CONSTRUCTION 791 


is somewhat larger than for steel construction, because the reinforced 
concrete construction is simpler and therefore results in a smaller 
total cost. If the headroom is not large enough for a concrete girder, 
steel girders must be used. In such cases the supporting columns 
below the girders also should be of structural steel. Sometimes the 
depth of girders to carry the offset columns is made equal to the 
story height and the girders are placed between the second and third 
floors. 

The center line of the tracks should be at least 8 ft. from the face 
of the building. The top of the loading platforms should be 4 ft. 
above the top of the rails. 

Loading Platform for Trucks.—Each warehouse of any size 
must have provisions for loading merchandise on trucks and unloading 
it from them. The truck-loading platform is usually placed inside, 
in order to make the loading operations independent of weather con- 
ditions. It is often necessary to omit some columns in the driveway 
to facilitate the operation of trucks. This is expensive, as the upper 
columns must be carried on heavy girders. The level of the driveway 
should be 3 ft. 8 in. below the floor level, so as to bring the platform 
on the level with the truck floor. 

Elevators.—In modern warehouses, elevators are used to move 
goods up and down, and their location, number, and size are therefore 
of particular importance. Ifit isimpossible to ascertain ahead of time 
the number of elevators required, special provision should be made 
for installing additional ones in the future. This is done by providing 
the same framing as used in the elevator shaft, and covering the 
opening with a temporary removable slab. This expedient is often 
resorted to in cases where provision is made for additional stories. 
Until the upper floors are added, the space may be used for storage. 
With the increase in number of stories, the temporary slab is removed 
and additional elevators installed to provide for the increase in 
traffic. : 

Floor Finish for Warehouses.—Granolithic finish is commonly 
used in warehouses. When there is much heavy trucking and the 
main travel of trucks is along a well-defined lane, special hard finish 
is provided for the truck lanes. 


792 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


COLD-STORAGE WAREHOUSES 


The foregoing discussion of warehouse construction applies 
equally to cold-storage warehouses. ‘The difference in construction 
is only due to the use of insulators in the cold-storage warehouses. 

The insulator usually consists of lith or cork board 4 to 8 in. 
thick, depending upon the degree of insulation required. The floors 
are insulated by laying the insulator on the concrete in hot asphalt. 
A 3-in. concrete slab is placed on the top of the insulator. To be 
effective, the insulation must be continuous. Not only the floors but 
also the columns must be insulated. The exterior columns usually 
consist of two sections, one carrying the floor load and the other the 
brickwork. A space is left between the two columns, sufficient for the 
placing of the insulator. For details of Insulation see Engineering 
New-Record, July 24, 1924, p. 140. 

Live Load for Cold Storage.—Live load depends upon the weight 
of goods to be stored and may vary from 150 to 250 lb. per sq. ft. 
Floors carrying refrigerator machinery and tanks should be designed 
for special loads. The live load for floor sections carrying freezing 
tanks should be 350 lb. per sq ft. To the load should be added the 
weight of the refrigerating coils suspended from the ceiling. | 

Type of Floor Construction.—Flat slab is the most commonly 
used floor construction for cold-storage warehouses. The state- 
ments made in regard to spacing of columns and other matters, in 
connection with ordinary warehouses, apply with equal force to 
cold-storage warehouses. 


MANUFACTURING BUILDIN GS 


Manufacturing buildings are of two types: One-story buildings, 
and multi-story buildings. One-story buildings are generally used 
where the process of manufacture requires breadth, length, and 
height. The roof spans in such buildings are large. 

While reinforced concrete 1s often used in one-story construc- 
tions (see Garfield Foundry, p. 666), steel truss construction will be 
cheaper in first cost. The additional cost of upkeep, however, may 
more than balance the difference in first cost. : 

Fireproof multi-story buildings are almost universally built of 
reinforced concrete. 

Spacing of Columns.—In designing a factory, a machinery layout 
of the floors should first be made, with due consideration of economy 


MANUFACTURING BUILDINGS 793 


of construction of the building. If no special arrangement of columns 
is required by the process of manufacturing, their spacing is governed 
entirely by economy of construction. While 18 to 20-ft. panels are 
usually most economical, larger spans may prove much more satis- 
factory from the standpoint of usefulness of the building. The addi- 
tional cost of long spans may be more than offset by greater freedom 
in the arrangement of the machinery. 

For buildings with special machinery, such as printing presses, 
the spans are governed by the dimensions of the panels. Thus, in 
the Youth’s Companion Building,” the floor plan for which is shown 
in Fig. 281, p. 794, the spans were governed by the arrangement of 
the presses. 

In shoe factories, where the process of manufacture progresses 
longitudinally with the building, construction two spans wide is con- 
sidered advantageous. Thus, in the Ideal Shoe Factory,? shown in 
Fig. 282, the spacing of the columns across the building is 20 ft. 

_ From the standpoint of cost of the structure, it is advisable to use 
equal spans throughout the building. However, to fit special require- 
ments, the dimensions of the spans may have to be varied. Thus, 
in the Clark Biscuit Co. building, the spacing of the columns across 
the building was not equal, as it was adjusted to suit the require- 
ments of the manufacturing process. 

Type of Construction.—For single-story buildings, the roof con- 
struction may be of any of the long-span roof types described on 
pp. 661 to 679. 

For multi-story buildings, flat slab construction is most com- 
monly used. In cases where the spacing of columns is not adaptable 
to flat slab construction, beam and girder construction may be used 
for heavy loads and light weight construction for light loads. 

Design Live Loads.—Design live loads depend upon the purpose 
for which the building is intended. Light manufacturing buildings 
may be designed for 100 to 125 lb. per sq. ft. Ordinary manufac- 
turing requires a live load of 150 lb. per sq. ft. Heavy manufacturing 
buildings may have to be designed for 200 to 400 Ib. per sq. ft. Por- 
tions of the building used for storage should be designed accordingly. 

Full live load should be used in designing beams and slabs, also 
joists in light-weight floors. For girders carrying an area over 300 


2 Densmore, LeClear & Robbins, Architects, Sanford E. Thompson, Consult- 
ing Engineer. 
3 Mowll & Rand, Architects, S.M.I. Engineering Co., Engineers, 


794. REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


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AUTOMOBILE BUILDINGS AND GARAGES 795 


sq. ft., as well as for flat slab construction, a reduction in live load of 
15 per cent is permissible. 


Loading Platforms.—See corresponding section on previous Bee 
under ‘‘ Warehouse Construction.” 

Floor Finish.— Because of the large variety of structures embraced 
in this section, no general recommendation as to floor finish is possible. 
Obviously, heavy manufacturing buildings require different treat- 
ment from light manufacturing buildings. In buildings with con- 
siderable trucking special floor finish may be required. 







































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Granolithic finish seems to be the best all-round type suitable 
not’ only for ordinary use but also for trucking. However, the oper- 
atives sometimes object to it, and for this reason hardwood floor is 
often used. In some processes, the slight dusting which occurs in new 
granolithic floors is objectionable. This dusting will occur even 
with the best granolithic, unless it has a ground surface. For further 
discussion of floor finish, see p. 620. 


AUTOMOBILE BUILDINGS AND GARAGES 


Concrete is used to a great extent in the construction of buildings 
for use in connection with automobiles. These may be divided into 
buildings for the manufacture of automobiles, which are similar in 


796 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


design to other manufacturing buildings, and buildings for the storage 
of cars, such as garages and service stations. The garages may be 
designed either as commercial garages, where the automobiles are 
handled within the building by garage men, OF as private garages 
where the owner drives his cars in and out. Such garages are built 
by hotels and department stores for their patrons. The difference 
sn the two cases in the arrangement of the columns may be consider- 
able. Buildings for service stations and storage will be discussed below. 

Live Load.—All garages and service stations should be designed 
for a live load of 150 lb. per sq. ft. This live load should be increased 
if the garage is likely to be used for storage of heavy trucks. Full 
live load should be used in the design of beam and girder or joist 
construction, for slabs, beams, and joists. For the design of the 
cirders, however, when the area carried by the girder exceeds 400 sq. 
ft., the live load may be reduced by 25 per cent. The same reduc- 
tion may be made in design of flat slab construction. The reason 
for the suggested reduction 1s obvious. While it is possible to place a 
wheel load on beams and slabs so as to vet the effect of the full live 
load, it is hardly possible to load a large area sufficiently to produce 
the effect of a uniformly distributed load of 150 lb. per sq. ft. 

Size of Cars and Trucks.—The spacing of columns and the width 
of aisles depends upon the overall dimensions of the cars. These 
are given below. 

Size of Passenger Cars.—In the table below, the various makes of 
passenger cars are divided into five groups, and the average dimen- 
sions for each group are given. 


Overall Dimensions of Passenger Cars 

















By re Width | Height 

Ford, Star, Overland, Chevrolet. £0 eee = 11 ft. 7in.| 5 ft. 5 in. | 6 ft. 4m. 
Durant, Essex, Buick, Studebaker, Maxwell, 

Dodge, Hupmobile, Dort......---++---> 13 ft. 6 in.| 5ft. 5 in. | 6 ft. 4m. 
Willys-Knight, Jewett, Franklin, Buick, 

Oldsmobile, Lexington, H. C. Stutz, . 

Hudson. cere ae ee ee ree 14 ft. 6 in.| 5 ft. 8 in. | 6 ft. 6 im. 
Nash, Apperson, Jordan, Paige, Marmon, 

Studebaker, Haynes. ...-----+-+s+ 7070? 15 ft. Gin. | Dito meO eee in. 


Cadillac, McFarlan, Pierce Arrow, Packard. | 16 ft. 6 in. | 5 ft. 8 in. | 6 ft. 7 in. 


TE 


In each group, cars are arranged according to size, 


AUTOMOBILE BUILDINGS AND GARAGES 


(ey 


From the above table, it is evident that an average space of 6 ft. 
6 in. by 15 ft. is sufficient for most makes of passenger cars. The 
width of the storage aisles, from center to center of columns, there- 
fore, may be made from 13 ft. 6 in. to 15 ft. 
size of Trucks.—The dimensions of trucks vary more widely 


than the dimensions of passenger cars. 


average overall dimensions for trucks of different capacities. 


Overall Dimensions of Trucks 


In the table below are given 


a a a ea ee 








Capacity in Tons Length Width Height 
5 to 7 15 ft. 5 ft. 6 in. ¢ ft, 10 in. 
1 to 13 18 tt. 5 ft. 9 in. 8 ft. 6 in. 
2 18 ft. 6 in. 5 ft. 9 in. 9 ft. 3 in. 
3 20 ft. 8 in. 6 ft. 6 in. 10 ft. Oin. 
5 21 ft. O in. 7 ft. 6 in. 11 ft. 9. 0<in; 





To accommodate trucks with high bodies, it may be advisable 
to make the story height of the first and second floors higher than 
for the rest of the garage. 

The Ramp Buildings Corporation estimate that 50 per cent of 
trucks in use are delivery cars no larger than passenger cars and may 
be accommodated in a space 15 ft. by 6 ft. 6 in. Forty per cent of 
trucks are of less than 2-ton capacity, requiring an average space of 
18 ft. by 7 ft.. The balance of trucks (about 10 per cent) require a 
space of 20 ft. by 7 ft. 6 in. to 8 ft. 

Spacing of Columns.—The spacing of columns should be equal to 
a multiple of the spacing required by a car, plus the maximum width 
of the column. From the table on p. 796, it is evident that the width 
of the car is from 5 ft. 5 in. to 5 ft. 8 in., or an average of 5.5 ft. 
Allowing 1 ft. between cars, the space required by one car becomes 
6.5 ft. This space is ample for average drivers. In some cases, in 
order to provide for less experienced drivers, a space of 7 ft. is pro- 
vided for each car. Larger space than 7 ft. is wasteful. 

If 6.5 ft. is accepted as the required width for a car and the max- 
imum width of the column is 2 ft., the required longitudinal spacing 
of columns for two cars per panel is 2X6.5+2=15 ft.; for three cars 
per panel, 3X6.5+-2=21.5 ft.; and for four cars 28 ft. Any spacing 
of columns between 15 and 21.5 ft. would not be economical as it 
would be too large for two cars but not large enough for three cars. 


798 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


In commercial garages, serving for storage of all makes of cars, the 
spacing of columns 1s made the same in all directions, as such garages 
are usually packed to capacity with cars, with comparatively small 
spaces for aisles and no clear passage from each car to the elevator 
or other means of exit. When it is necessary to get the car out, the 
cars standing in its way are shifted by the attendants. 

In garages where the cars are brought in and out by the owner, 
special spaces are provided for storage of cars and separate aisles for 
getting in and out. These aisles are usually kept clear and have a 
direct connection with elevators or ramps. The columns are then 
arranged with a wide lateral spacing in the center, forming the aisle, 
with two smaller spacings of columns, one on each side of the center 
aisle, used as storage space. This group of three rows of panels is 
repeated all through the floor (see Fig. 284, p. 800). The widths of 
the aisle and the two side spans depend upon the size of the cars to 
be stored. In garages used for general storage, it 1s necessary to 
make provision for the largest car likely to be stored. A spacing of 
columns of 23 ft. 6 in. to 25 ft. for the center aisle and 13 to 15 ft. 
for the side spans is considered sufficient. ‘The longitudinal spacing 
of the columns should be arranged so that the space can be used 
either for two or three cars. The spacing required for two cars is 
15 ft. and for three cars 22 ft., allowing 6.5 ft. per car and 2 ft. for 
maximum size of the column. 7 

Ramps.—Garages are usually provided with ramps leading from 
the street to the first story and also to the basement. To make the. 
descent into the basement less steep, and also to provide ventilation 
for the basement, the first floor is placed several feet above the ground 
level, In garages consisting of several stories, the cars are usually 
brought up and down in elevators. In many instances, however, 
ramps are provided to all floors and the automobiles are brought 
up and down under their own power. 

Details of Ramps.—The grade of ramps may be made as large 
as 15 degrees. The surface of the motor ramp should have a wood- 
float finish. Ramps near the street entrance should be treated so as 
to prevent the rear wheels of the cars from slipping when the ramps 
become wet. A very economical and satisfactory plan is to have 
them grooved every 4 in. with transverse ““V” grooves, 7 in. to 
1 in, deep, depending upon the thickness of the slab. Curbs 8 in. 
wide and 10 in. high should be provided on both sides of the ramp 
and also heavy pipe rails. The curb should wrap around columns 


799 


AUTOMOBILE BUILDINGS AND GARAGES 


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to prevent mud guards from touching them. The view of both 
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of collision between the cars coming in and those going out. 


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3k (9.3 e- 226" ake - 22-6 “4 ke 2216" He 8.6 TE DO 


Ad, 
Plea isk | moe 18) 198" nanan names enc tansts 49'S 


Fig. 284.—Details of Statler Garage in Buffalo. (See p. 801.) 
Geo. B. Post & Son, Architects. United Fireproofing Co. Contractors. 


A curved inclined driveway, as used in the six-story Washington 
Garage + in Chicago, is shown in Fig. 283, p. 799. The average” 


‘Holabird & Roche, Architects. For further description see Engineering 
News-Record, July 24, 1919, p. 188, . 





AUTOMOBILE BUILDINGS AND GARAGES 801 


inclination of the driveway is 11 degrees, with 13 degrees at the 
inner curve and 9 degrees at the outer curve. The width of the 
driveway between curbs is 10.ft. 3 in. on the tangents and is increased 
to 11 ft. on the turns and reduced to 8 ft. at the middle of the curve. 

D’Hume Ramp System.—With ordinary height between floors, 
the ramp must be very long and therefore must occupy too much of 
the garage space. This is remedied in the D’Hume Ramp System, 
by constructing the garage in two parts so arranged that the story 
in one part falls halfway between the stories in the other parts. 
With this arrangement it is necessary to build ramps only large 


8 Concrete Cap 
V.- 12" Brickwork 






Steel Sash 
/4 420" rine 


level /4 








level 13 






Pipe Rail... 
8x8 je 





I nr We 









level 8 


2a 






Fig. 285.—Section through D’Hume Ramp. (See p. 801.) 


enough for one-half of the story-height. A design containing 
‘D’Hume ramps is shown in Fig. 284, representing a floor plan and 
cross section of the Statler garage in Buffalo. Photograph of the 
exterior of this building is shown in Fig. 262, p. 750. It may be of 
interest to mention that in this building slag was used as coarse agere- 
gate for concrete. Section through the ramp is shown in Fig. 285, 
p. 801. . 

Circular Ramps.—Another solution of the same problem is 
employed in the Eliot Sheet Garage in Boston, where a continuous 
circular ramp is provided. This arrangement is less satisfactory than 
the previous one, as it occupies too much space. 


802 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


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(‘208 ‘d aag) “sseyA[ “WOJSOg Ul UON}BIS ZdIALIG UlP{UVIY— "98S “Ol 














J0opy PIE OF a pune YSDpl 


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1009 Pd/2D ULL. 
aids] 100] UO 7,9 \ ‘ suowips0d DL ,P- 





Jof saauan youg Ny 


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OFFICE BUILDINGS 803 


Typical Arrangement of Columns.—Figure 286, p. 802, shows the 
arrangement of columns as used in Franklin Service Station in 
Boston, designed by Shepard & Stearns, Architects, 8. M. I. En- 
gineering Co., Consulting Engineers. The center span is 30 ft. wide, 
while the end span, on both sides, is 21 ft. 4 in. wide. The longi- 
tudinal spacing of columns is 27 ft. 93 in. 

Construction of Garages.—Flat slab is the most widely used floor 
construction for garages. It is used for spacings of columns up to 
30 it. 

In cases where the spacing of columns is not well adapted to flat 
slab construction, beam and girder or light-weight floor construction 
may be used. The beam and girder or clay tile construction is prefer- 
able to the joist construction, because the thin slab between the 
joists is not well adapted for carrying heavy concentrated loads, 
unless provided with special reinforcement. Also, the thin slabs 
and narrow exposed joists would suffer considerably in case of a fire. 

Floor Finish.—Granolithic floor finish is practically always used 
in buildings intended to house automobiles. 


OFFICE BUILDINGS 


Reinforced concrete is being used in office buildings to an increas- 
ing extent. Ordinarily, concrete is cheaper than fireproofed struc- 
tural steel, particularly when flat ceilings are required. 

Limit of Height for Concrete Buildings.—Contrary to popular 
opinion, there is no limit to the possible height for concrete buildings, 
except that imposed in each particular case by considerations of 
economy. While not long ago a height of twelve stories was consid- 
ered as the practical limit, even by engineers engaged in concrete 
construction, in recent years a number of buildings of considerably 
greater height have been erected. Usually, there is an objection to 
tall concrete buildings on account of the large size of the columns 
required in the lower floors. This objection is easily removed by sub- 
stituting, in these floors, structural steel columns encased in con- 
crete. 

The Medical Arts Building 5 at Dallas, Texas, is a notable exam- 
ple of a tall concrete building. It is 19 stories and 255 ft. high above 
basement floor. 

°C. E. Barglebaugh, Architect. For description see Concrete Frame and 
Exterior for High Tower Building, Engineering News-Record, April 5, 1923, 
p. 610. 


804 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


The Hide and Leather Building, in New York,® is 18 stories, 215 ft. 
above the curb. In both of these instances, the concrete frame 
proved cheaper than structural steel. 

Shape of Building.—The most important problems in the design 
of office buildings are those of light and ventilation. In modern 
office buildings, the courts are arranged so as to limit the width of 
the building to 75 ft. A typical floor plan is shown in Fig. 287. 
The passenger elevators are placed in one central location instead of 
being divided into groups. 


J 


pox 














Fig. 287.—Typical Floor Plan for Office Building. (See p. 804.) 


Theoretically, in long buildings, two groups of elevators, one at 
each end of the building, would serve the purpose better. However, 
offices very often occupy large areas, extending the full width of the 
building. In such cases it is not possible to make all offices accessible 
from both groups of elevators, and confusion results, especially for 
visitors. 

Spacing of Columns.—In old office buildings, columns were placed 
along the sides of the central corridor. The cross section then con- 
sisted of large end panels and a small center panel. Such an arrange- 


6 Designed and built by Thompson & Binger Co., Raoul Gautier, Chief Engi- 
neer. 


HOTELS, APARTMENT HOUSES, AND HOSPITALS 805 


ment was satisfactory for small offices or subdivided offices. Now, 
with the advent of large, open office spaces, the office often occupies 
the full width of the building. Equal spacing of columns is therefore 
much more advantageous. This also results in cheaper construction, 
as the large spans required in the first case are replaced by smaller 
spans. If necessary, the space can be subdivided into smaller offices 
without difficulty. The corridor may be placed either in the center, 
dividing the space into offices of equal depth, or off center. 

Loading.—A live load of 70 lb. per sq. ft. is usually sufficient for 
the upper floors of an office building. The first floor is usually 
designed for 150 lb. per sq. ft. In some locations it is advisable to 
design the office building for 120 lb. per sq. ft., to make it available 
for light manufacturing. 

Dead load should include a proper allowance for partitions. This _ 
is usually 25 lb. per sq. ft. Weight of plaster should also be included 
in the dead load. 

In light-weight floors, special joists should be provided for heavy 
partitions running in the direction of the joists. To provide for par- 
titions by increasing the dead load is obviously insufficient. The 
partition load is carried by one joist only, as the thin topping is not 
sufficient to transfer the load to the adjoining joists. 

Types of Construction.—Light-weight floors, such as clay tile 
floors and metal tile floors with plaster ceiling, are considered the 
most economical for office buildings. 

Flat-slab construction is now gaining in favor, especially for 
office buildings designed for heavier loads. The original objection 
of architects to flared heads is gradually disappearing, especially 
since, with increased knowledge, the size of column heads and their 
shape may be made to suit architectural requirements. The interior 
of a flat slab office building is seen in Fig. 269, p. 761. 

Finish.—Hardwood finish in offices and terrazzo or granolithic 
finish in corridors have been used very extensively. In recent years, 
linoleum, or rubber parquetry, is growing in favor. Also, grano- 
lithic finish, usually painted, is often used, but objection is made to it 
on account of hardness, noise, and conductivity of heat. 


HOTELS, APARTMENT HOUSES, AND HOSPITALS 


Reinforced concrete is well adapted for the construction of hotels 
and apartment houses, In hotels, a corridor is usually placed in the 


806 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


middle and the rooms on both sides of it. In old hotels, the columns 
were placed in the corridor wall, so that the width of the building 
was divided into two large spans at the ends with a short span in the 
middle. In modern hotels, each room has a bathroom or closets 
placed next to the corridor. This makes it possible to place the 
columns in the corners of the bathroom, farther away from the cor- 
ridor, so that the end spans become practically equal in length to the 
center span. The cost of construction of the equal spans is smaller 
than that of the spans used in the old arrangement. The longitudinal 
spacing of the columns is governed by the size of the rooms. Similar 
construction is possible in apartment houses. 

In hospitals, it 1s usually necessary to place the columns in the 
corridor wall. 

Design Loads.—Hotels, apartment houses, and hospitals should 
be designed for a live load of 50 Ib. per sq. ft. for private rooms and 
100 lb. per sq. ft. for corridors and public rooms. In computing the 
dead load, proper allowance should be made for partitions. The 
weight of plaster and flooring should also be added to the dead load. 
These items are often neglected, with the result that the design is of 
inadequate strength. 

Type of Floor Construction.—The economical type of floor con- 
struction naturally will vary with conditions.. Usually, light-weight 
construction, as described on pp. 588 to 611, will prove satisfactory. 

Flat slab construction is not used to any great extent, because of 
the objection to the flared heads. However, for spans up {O1LS- ih, 
it is possible to build flat slabs economically without flared heads. 
In many cases, the rooms can be arranged so as to permit even spacing 
of columns not more than 18 ft. apart, which permits the use of flat 
slab construction (without flared heads), at a cost much lower than 
that of light-weight slab. This is not possible where spans over 18 ft. 
are required in the bulk of the structure. 

Floor Finish.—The most popular floor finish for this type of build- 
ing is hardwood flooring, particularly in the living portions. Ter- 
razo is generally used for corridors, kitchens, service portions and 
bathrooms. | 

In recent years, linoleum and similar coverings are gaining in 
favor on account of their lower cost and great durability. With 
linoleum, there is also a gain in thickness of about 2 in. 

Objections are made to the use of granolithic finish for buildings 
of this type. The hard appearance, which constitutes one of the 


HOTELS, APARTMENT HOUSES, AND HOSPITALS 807 

















pee 
~ < Clinie 
Laboratory 
Mm _-4-Surgical 
Students mil oat lass ates 
arto! Coat Room. |_| | <0 
2 Special Clinic | i 
* Laboratory~ 7 e 
& = Internes-~.. 





Solarium 


F iq. 288.—Layout of Rooms in Hospital for State of Wisconsin. (See p. 808.) 





North 





Win 7 













laced _voists. 
“21 
_y_ EEA re 3t of \4 SSS II, 
Lnteys 7-I"Rods in top 













/4-I"Rods in bottom Section A-A 


re ge! “UPS Detail at X-Y 
& Rod, 1749" 2 Rod, 22-3" 


eee eee | 5 
2 Rod,4' 


35 Joists 


| 

























Dumb waiter shaft 





















9 S3 joist, Interior ¢ of --7, 

Et SP ous GF tiers columns building 

il j 

BiJoists . | 

34 Rows of tile | loggia , 

iia? y Wall 4 Ay tee/ columns | 
Les Oz columns Beam Ib JUUL in loagia os 


ae 
Section of Column X? — S-Lintel Z 
Lintels Z 







2" Rod! 
Spe gn 


Section o 
other Lintels 


Linte/] Z 


Fra. 289.—Design of Floor Construction for Hospital for State of Wisconsin. 
(See p. 808.) 


808 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


objections, can be remedied by painting the floors a warm 
color. 


Design of Hospital—A typical design of a hospital for the State 
of Wisconsin” is shown in Figs. 288 and 289, p. 807. The arrange- 


Third Floor Gr. 61.60 


ee =e ee ee ee 


First Floor Gr. 29.60 





Fic. 290.—Typical Cross Section Showing Classrooms Opening on Corridor. 
(See p. 810.) 
Massachusetts Institute of Technology, Cambridge, Mass. 


ment of rooms is evident from Fig. 288, while the construction details. 
are shown in Fig. 289. 


SCHOOL BUILDINGS 


Reinforced concrete is used in the construction of school buildings, 
either for the whole frame or for floors only. 

Live Load.—The following live load may be safely used in the 
design of the various parts of the school building: 


7Structure and Facilities of a Modern Hospital. By Arthur Peabody, 
Engineering News-Record, April 10, 1924, p. 608. 


SCHOOL BUILDINGS 809 


Live load lb. 


per sq. ft. 
Bee ey ne a ae 50 
Pea ANC SUAITS: oo. ee one ee ke ns 100 
Assembly rooms and auditoriums....... 100 
Oo) ToL! ES Sel lead a aaa a 100 
ee i i gs bec ot an 100 


Wall-bearing vs. Skeleton Construction.—School buildings are 
seldom over three stories high; therefore, wall-bearing construction is 
used to a great extent. Recent investigation show that skeleton con- 
struction may be built at the same cost and sometimes cheaper. It 
has the additional advantage that the concrete structure may be 
erected without any interference from the brickwork. 

Type of Floor Construction.—The type of floor construction 
depends upon the size of the room and the purpose for which it 
serves. 

Class Rooms.—The average class rooms are 32 ft. long and 22 ft. 
wide. They are arranged either on both sides of the corridor or on 
one side only. In wall-bearing construction, the walls along the 
corridor may be built as bearing walls. The floor construction then 
spans from wall to wall. The cost of the construction may be 
reduced by omitting the interior bearing walls and substituting 
columns placed along the corridor, with longitudinal beams to carry 
the floor as well as the corridor partitions. The class rooms must 
be built without any interior columns. Flat ceiling is also often 
required. 

In class rooms beam and slab construction may be used for the 
floor, the beams spanning across the building, which is the short direc- 
tion of the rectangle. The spacing of the beams should be about 
8 ft., which permits the use of 3}-in. slab. When flat ceilings are 
required, a suspended ceiling, consisting of steel rib lath suspended at 
the beam and at one or several intermediate lines, should be used. 

Where hollow tiles or steel tiles are available, light-weight con- 
struction, as described on p. 588, may be used for the floor. The 
joist may span in the short direction only, or a two-way system may 
be used, with joists running in two directions at right angles to each 
other. . 

Sometimes a solid concrete slab, reinforced in two directions, 
may be used. 

The hollow tile and solid concrete ceiling may be plastered directly 


810 REINFORCED CONCRETE IN DIFFERENT TYPES OF BUILDINGS 


below the construction with two-coat plaster. Metal tile construc- 
tion requires metal lath underneath with three-coat plaster. 

Special Rooms.—In ‘ndustrial training schools, the sections 
occupied by workrooms, shops, and drawing-rooms are much larger 
than the class rooms, but there is no objection to intermediate col- 
umns. Columns may be spaced with due regard to the economy of 
the structure. Flat slab construction is often used in special rooms. 

Gymnasiums and Auditoriums.—Rooms serving for gymnasiums 
‘and auditoriums are usually of considerable length and width, with- 
out any intermediate columns. 3 

In some cases, an attempt Is made to combine the gymnasium 
and auditorium. This is usually not successful, because the floor of 
a gymnasium must be level and therefore is not well adapted for 
auditorium purposes. 

Gymnasium and auditorium are often built side by side, so that 
the gymnasium or part of it may serve as a stage for the auditorium. 
In such case, means are provided for shutting off the whole or a part 
of the gymnasium so that both rooms may be used separately, if 
desired. 

Usually, no additional story is placed above either of these rooms. 
The roof construction is ordinarily carried by steel trusses. While 
these are cheaper as to first cost than concrete members, they require 
larger cost, of maintenance. If a permanent roof is required, it may 
be built as explained in connection with long-span roof construction 
(p. 661). 7 

The galleries and the ‘nclined floor in the auditorium may be 
designed as explained in connection with theater construction. 

Building for Massachusetts Institute of Technology.—Figure 290, 
p. 808, shows typical framing plan for Buildings 1, 3 and 5 of the 
Massachusetts Institute of Technology The beams were arranged 
across the building. In portions of the building where classrooms 
are placed on both sides of the corridor, the columns are placed along 
the corridor. Fig. 290 gives the widths of the classrooms and of the 
corridor. 

Stairs.—As all the up-and-down travel in school buildings is by 
means of the stairs, their number and location must be carefully 
worked out. The stairs must also serve as fire exits and must be 


8 Designed and built by Stone & Webster Engineering Corporation; William 
W. Bosworth, Architect. The concrete design and construction was under the 
supervision of Sanford E, Thompson, Consulting Engineer, 


SCHOOL BUILDINGS 811 


properly protected. As a general rule, each school building should 
have at least two staircases, located at opposite ends of the corridor. 
A larger number may be required. 

The width of stairs is usually 4 ft. 6 in. The rise of steps varies, 
according to the age of the pupils, from 6} to 7 in., and the tread 
from 104 to 11 in. 

Floor Finish.—The floor finish in class rooms, auditorium, and 
gymnasium is usually of hardwood. Linoleum and rubberized par- 
queting also give satisfactory results. 

For corridors, lavatories, and other service portions, granolithic 
finish or terrazzo may be used. 


CHAPTER XX 
REINFORCED CONCRETE CHIMNEYS 


FORMULAS FOR REINFORCED CONCRETE CHIMNEY AND 
HOLLOW CIRCULAR BEAM DESIGNS 


REINFORCED concrete chimneys may be regarded as vertical can- 
tilever beams supported at the base. The loadings to be provided 
for are (1) the weight of the chimney and (2) the wind pressure. 
Although the design is somewhat complicated by the fact that the 
beam is circular and hollow, the treatment is nearly identical with 
that of ordinary rectangular beams. In fact, the analysis which 
follows is based upon the several fundamental assumptions adopted 
in reinforced concrete beam design with only one additional assump- 
tion, viz.: that, since the concrete is usually thin as compared with 
the diameter of the chimney, no appreciable error 1s involved in 
assuming all material as concentrated on the mean circumference of 
the shell. An analysis for shear 1s given on p. 819. An example of 
chimney design and review is given on p. 829. 

Although specially devised for a chimney, the formulas are appli- 
cable to any hollow beam. 

The principles involved in the demonstration of the thickness of 
steel and concrete are taken by permission from the analysis by 
Messrs. C. Percy Taylor, Charles Glenday, and Oscar Faber.! 


NOTATION 


W = weight of the chimney above the section under consideration, 
lb. ; 
M = bending moment about that section due to wind or other cause, | 
in.-lb.; 
P = total compression in concrete, Ibs 
T = total tension in steel, lb.; 
n= i — ratio of modulus of elasticity of steel to that of concrete ; 
1 Engineering (London), Mar. 13, 1908. | 
812 


DERIVATION OF CHIMNEY FORMULAS 813 


fc = maximum compression unit stress in concrete (measured at the 
mean circumference), lb. per sq. in.; 
fs = maximum tevsion in the steel, lb. ate Sq. in; 
D = mean diameter of shell, in.; 
r = mean radius of shell, in.; 
¢ = total thickness of shell, in.; 
te = thickness of concrete ailyei in. 
ts = thickness of an imaginary steel shell of mean radius 7, and 
having a cross-sectional area equivalent to the total area of 
reinforcing bars, in.; 
total area of vertical reinforcing bargin the section under 
consideration, sq. in.; 
k = ratio of distance of neutral axis, from mean circumference on 
compression side, to diameter D; 
j, 2, Cp and Cp = constants for any given value of k. (Tables 1 and 
2, p. 828); 
jD = distance between center of compression and center of tension; 
2D = distance from center of compression to center of force due to 
weight. 


= 
T 


DERIVATION OF CHIMNEY. FoRMULAS 


Referring to Fig. 291, if f; is the maximum. intensity of stress in 
the concrete at the mean circumference on the compression side, then 
the intensity of compression in the steel 
at that point is nf,. Since f, is the 
maximum intensity of stress in the 
steel at the mean circumference on 
the tension side, then the variation 
of the stress in the steel, across the 
section cd, is represented by the 
straight line ab, which cuts the line 
cd at e, thus locating the neutral axis 
or the line of zero stress. A con- 
stant value having been assumed kf > 
for the modulus of elasticity of the Fy. 291.—Resisting Forces in a 
concrete in compression, it therefore Reinforced Chimney. (See p. 813.) 
follows that, at any point of a given 
section, the stress in either the concrete or the steel is directly pro- 
portional to the distance between that point and the neutral axis. 





LLM 


Neutral 


_ Center _ 





ont dit ea bae 


814 REINFORCED CONCRETE CHIM NEYS 


Calling kD the distance between the neutral axis and the mean 
circumference on compression side, as shown in Fig. 291, we have by 
similar triangles 





kD __h 
D) topahe ete 
whence 
1 
a i 
1 
Nc 


By this formula, the position of the neutral axis may be determined 
for any combination of fe, fs, and n. The formula for k is the same 
as for rectangular beam. 

If now, as shown in Fig. 292, represents half the angle subtended 
at the center by the portion in compression, we have 


cos a = (1 — 2k) 


from which, for any given value of k, cos becomes known, as well 
asa and sina. Thus, having located the neutral axis for any given 
etnias combinations of f.,/,and , and 
ra ie 7 ~Centerof bearing in mind that the stress 
ae las at any point of the shell is pro- 
portional to the distance be- 
tween that point and _ the 
neutral axis, we can now de- 
termine the total force on the 
ei compression side, the total 
Tension force on the tension side, and 
also the location of the center 
of compression and the center 
of tension. 
Considering a small radial element subtending an angle dé, as 
shown in Fig. 292, we have in this element, since the length of an 
arc is its radius times the angle, 













Center of f. 
Compression --_f-/f., 


Concrete in---%¥X\1 
Compression 
Steel in -~ 


Compression 


Fic. 292.—Distribution of Stresses in a 
Reinforced Concrete Chimney. (See p. 814.) 


area of concrete = t.7rdé, 
area of steel =Hrdu: 


The distance from the element to the neutral axis is r(cos 6 — cos a), 
while the distance from the neutral axis to the point of extreme stress, 


DERIVATION OF CHIMNEY FORMULAS 815 


fc, is r(1 — cosa). Therefore, the intensity of stress on this ele- 
mental area is 


EAGOS =: C08 cr) in the concrete, 
r{(1’— cos a) 


te 





and 


r(cos 8 — cosa) . 
te Perce) in the steel. 





Assuming these intensities at the mean circumference to represent 
the average for the entire element, we have the total force on the 
elemental area (concrete and steel) 





cs f.r(cos 6 — cos a) 
dP = (t, + nt,)rdé Milne. 


The total compression force, P, acting across two alpha is, therefore, 


“fer(cos 8 — cos a) 


P= (to + nt,)2 , (1 — cos a) 


dé. 





Integrating this expression, we obtain 
2 : 
Piss Werth + Ms) eos gy cin Qa — a cos a). 


Since any given position of the neutral axis determines a, as shown 
above, this equation may take the form 


LP = 10rtiee nt iB ieiehanaadn i avons leah gn 


in which Cp is a constant for a given position of the neutral axis. 
(See Table 1, p. 828.) 

When the magnitude of P has been determined, its location, with 
respect to the neutral axis, may best be found by taking its moment 
about that axis and dividing by P, thus giving the distance from the 
neutral axis to the center of compression 1,, as shown in Fig. 292. 

As before, the compressive force on an elemental area is 


oe f.r(cos 6 — cos a) 
The distance of this force from the neutral axis being r(cos 6 — cos a), 
we have as its moment about that axis 


fr?(cos 86 — cos a)? 


dM, = (t, + nt.)rdé race awe 


>) 


816 REINFORCED CONCRETE CHIMNEYS 


while the moment of the total compressive force P is 


ate + niga f r( GOS Bare cos a)? f.r(cos 8 — Cos a)" 7, 


(1 = cos "(1 0os poe 


= (t, + nts) ee afer | (cos? 6d 


cos a) | Jo 


— Zeosa f cos 6d0 + cos? af as}. 
<Z005 ; ; ‘ 0 


Integrating, we have | 








M, = (te + nts)fr? al 


Sere aye cos? a. — § SiN a COS @ + 3a). 


Dividing M. by P, we have 


M. (w cos? a — $sin a cos a + 30) 


lL = = Tste> es Te (2) 


P (sin a — a COS @) 








Following a similar method of procedure, it is possible to deter- 
mine the total tension and the location of the center of tension. 

In accordance with our assumption that the concrete does not 
resist any tensile stress, it 1s evident that in considering the forces on 
the tension side of the section we are concerned merely with the steel. 
On the tension side, a small element, therefore, has an area = t.rdé. 

The intensity of stress on this element, being proportional to its 
distance from the neutral axis, 1s 


r(cos 8 + cos a) — 


Js r(1 + cos a) 
while ve total tension on the small element is 


(cos @ + cos a) 
(1 + cos @) 


The total force, 7’, on the tension side of the section is, therefore, 


(x — a) 
T =2 ip dip W008 23 ane a) 49 
0 





av Soar el ot 





(1 + cos a) 


Integrating, we have 





T = Farley (sin a + (t — @) COs @). 


+ cos a) 


DERIVATION OF CHIMNEY FORMULAS 817 


Since, as before, any given position of the neutral axis determines a, 
this equation may take the form 
OSCE ERS OREM SRE at 


in which Cy is a constant for a given position of the neutral axis (see 
Table 1, p. 828). By a method similar to that used in considering 
the force on the compression side, we may write the moment, about 
the neutral axis, of the force on a small element on the tension side as 


r(cos 6 + cos a)? 


(L+ cosa) ’ 
while the moment of the total tensile force, 7, about this axis is 


(7 — a) . vs 
ab: r(cos 6 + cos a) 
ea af SO ayaa eyede! 





dM r= t.rd of. 


Integrating, we have 
2 : 
AD cy eon anes ids = 2 3 ae 
My = 1,7 Iq A he a) cos? a + 3 sina cosa + 3(r — a)]. 


Dividing M, by T, we have as the distance of the center of tension 
from the neutral axis | 
( (7 — a) cos? a + 2 sina cosa + 1(r — a) ) 
- (sina + (x — a) cosa) 





ls = eet (4) 
From Formulas (2) and (4), it is evident that the distance be- 
tween the total force in compression and the total force in tension 
(i.e., 11 + 2) may, for any given po- 
sition of the neutral axis, be expressed 
as a constant times the diameter D. 
Thus, 1; + lz = jD, as shown in Fig. 
293. Likewise, as shown in Fig. 293, 
zD may represent the distance from 
the center of compression to the 
center of the chimney, z also being 
a constant for any given position of 

the neutral axis. 

In a chimney, the tensile and Fie. 293.—External and Internal 
compressive stresses which we have Forces Acting upon a Chimney. 
been considering are produced by (See p. 817.) 

a combination of wind pressure 
and the weight of the chimney. Thus, on any horizontal section, 
cd, as shown in Fig. 293, the forces external to that section are: 


<—__. 
Wind 





818 REINFORCED CONCRETE CHIMNEYS 


the horizontal pressure of the wind, causing a moment M about the 
section, and a central vertical load W representing the weight of that 
portion of the chimney above the section under consideration. 
These forces are resisted, and held in equilibrium, by the forces P 
and T, which represent the compressive and tensile stresses in the 
concrete and steel. 

The system of forces, as shown in Fig. 293, must be in equilibrium. 
Hence, taking moments about the force P, we may write 


TID = M — WeD: 


But 
YA OO Ba 
Therefore 
CrfrtgD = M — W2D. 
Whence 
aes M — WzD 
: Cn Dae 
The total area of steel, A, = 277ts. 
Therefore 
ad In(M — W2zD) 
fA are CfgD GAT haga Se ity sae 


From Table I, p. 828, it may be seen that the constant 7 changes 
but slightly for a considerable variation in the position of the neutral 


axis. Taking * = § for all cases, equation (5) may be 


_ 8(M — WeD) 
As = =o37 7 ae (6) 








While this formula is not exact, the error involved is inappreciable 
for almost any case, so that formula (6) may always be used in- 
stead of formula (5). 

Applying now the condition that the summation of all vertical 
forces must be zero, we have 


P—-T=W. 


Substituting values of P and T as previously found, the equation 
becomes | 

Crfar(te + nts) — Cafert, = W. 
Transposing and solving for ¢., we obtain 


ba W a (Cof — Cofen)rts 


te G pf c¥ 


DERIVATION OF CHIMNEY FORMULAS 819 
The total thickness of the shell is 
t=t,+4,, 


gk W oh (Cofs eT Cpfn)rt, 
Cpf r 


For convenience in use, after A, has been determined by the 


whence 


= ee 


formula given above, by substituting r = = and t, = ab this formula 
1. ; 


for t may best be written 


QW + (Crh — Caf.n)At 


t= A, 


Meee at +. ype ° . ° . (7) 


In view of the fact that Formulas (5), (6) and (7) contain the 
constants z, J, Cr and Cp, which, as has been shown, are dependent 
for their value solely upon the location of the neutral axis, it is evi- 
dent that, for any specific values of f., f,, and n, which in turn will 
determine the position of the neutral axis, the expressions for A, 
and ¢ will admit of a further simplification. For given values of 
Jc, fs and n, the necessary thickness of shell and area of reinforcement 
may be expressed merely in terms of the moment of the wind, M, 
the weight, W, and the mean diameter, D. The expressions, as 
given, however, seem best adapted to general use, and when sup- 
plemented by the tables given on page 828, are rendered quite sim- 
ple of solution for specific values. 

In Table 2, p. 828, are given values of &, the location of the neutral 
axis, for various combinations of f,, f, and n; while Table 1, p. 828, 
gives the corresponding values of the constants C p, Cr, 2 and j for 
various positions of the neutral axis. 

Shear, or Diagonal Tension—When the necessary thickness of 
shell and vertical reinforcement have been determined, the size and 
spacing of the circular steel hoops must be considered. The external 
forces produce shear and diagonal tension, which may be analyzed 
similarly to like stresses in rectangular beams, and the reinforcement 
necessary to resist the diagonal tension, which is a function of the 
vertical tension, may be determined. Usually, this reinforcement 
is not so great as that which it is advisable to insert for the proper 
distribution of temperature stresses, but nevertheless it should be 
determined, in order to be sure that it is sufficient in quantity. 

The concrete should never be relied upon to carry any tension or 


820 REINFORCED CONCRETE CHIMNEYS 


vertical shear, because the expansion from the heat may cause vertical 
cracks in the concrete. These need not be considered dangerous, if 
sufficient horizontal reinforcement is provided, any more than the 
vertical cracks in a brick or tile chimney. Considering the stresses 
due to vertical shear, it may be easily shown that at any horizontal 
section of a chimney the vertical shear per inch of height is the total 
horizontal shear on that section divided by the distance between 
centers of tension and compression, jD. With this as a basis, there 
may be developed a formula for practical use in determining the 
necessary area and spacing of horizontal steel hoops at any given 
section. Thus, 


Let i; = height of chimney above section under consideration, ft. ; 


F = effective wind pressure against chimney, lb. per sq. {t.; 
f, = allowable tensile stress in steel hoops, Ib. per sq. in.; 
D = mean diameter of shell in. ; 

po = ratio of area of steel hoop to area of concrete. 


The total external shear on any horizontal section of a chimney is 
equal to . 


D 
V — jah 


while the maximum longitudinal shear per inch of height is from 
beam formula 


‘ 
Die Sy: 

_DAF_ hk 

1X4) = 95 GD aoe 


Having seen that for all positions of the neutral axis 7 remains 
practically constant, and giving Jj an average value of, say, 0.783, 
we obtain, as the expression for the maximum vertical shear per inch 
of height, 

v X 2t = 0.106hF, 
while the shear or diagonal tension in one foot of height is 
Stresses to be resisted by hoops = 12 x 0.106/F. 


The area of steel in one foot of height of chimney will be 12¢po, 
and the stress the hoops in this height are capable of sustaining on | 
their two sections is 


Available strength of hoops = 2 X 12tpofs. | 


SLIM CHIMNEYS 821 


Equating the stress to be resisted with the available strength, we 
have | 


12 X 0.106h,F = 2 x 12tnof,, 
whence 
ehh ied 
WGA TS art 


This ratio of steel is for shear or diagonal tension only. To 
provide for temperature stresses, or rather to distribute the strains 
so as to prevent the localization of cracks, an additional amount of 
horizontal steel is needed. This may be provided for arbitrarily, 
by assuming 0.25 per cent steel, or rather 0.0025 for temperature 
stress in addition to the steel for shear. Expressing this as a formula 
for ratio of steel, we have 

h,F 


The use of small rods, spaced 6 to 10 in. apart, except in the upper 
part of the stack where the spacing may be greater, is advised. 

The spacing of hoops in many of the chimneys already built has 
been 18 to 36 in., but as such chimneys have frequently cracked 
quite seriously, more recent designs have called for 8- or 9-in. spacing 
throughout the entire stack. | 

Note on Slim Chimneys.—Since, in designing a chimney, the 
selection of certain allowable working stresses in the concrete and 
in the steel will fix the position of the neutral axis, it is evident that 
the ratio of these working stresses limits the compressive area of the 
section. Hence, for a very high chimney in which there is a large 

compression in the lower sections, it is possible that the selection of 
an ordinary working stress in the steel of 14.000 or 16 000 Ib. per 
Sq. in., together with the customary working stress in the concrete 
of, say, 500 Ib. per sq. in., would locate the neutral axis so near the 
compression side of the section as to make it impossible to obtain 
sufficient compression area to withstand the compressive forces with- 
out exceeding the allowable unit stress in the concrete. 

If, therefore, the thickness of shell as computed from Formula 
(7), p. 819, should work out materially larger than the assumed 
thickness, recomputation should be made on the basis of a smaller 
working stress in the steel, thus changing the position of the neutral 
axis so as to allow a larger proportion of the section to carry com- 
pression. In such a case it may be necessary to make a series of 


822 REINFORCED CONCRETE CHIM NEYS 


trials with different working stresses in the steel until the computed 
thickness checks with the assumed thickness. In high chimneys of 
small diameter, it may be impossible to utilize a working stress in the 
steel greater than 7 000 or 8000 Ib. per sq. in. 


DESIGN OF HOLLOW CIRCULAR BEAMS 


The analysis of a hollow circular reinforced concrete beam, whose 
thickness, compared with its diameter, is small, is similar in principle 
to that of a chimney. In this case the weight of the member acts in 
the same direction as the external forces, so that in Formulas (6) 
and (7), W, the weight in the axial direction, is zero. The forces of 
compression, P, and tension, 7’, are equal. The area of steel and the 


thickness of shell are therefore obtained from Formulas (6) and 
(7), pp. 818 and 819, by making W = 0. 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 


High factory chimneys of reinforced concrete are being built in 
this country and abroad. ‘The cost, especially of those over 100 ft. 
high, is usually much less than that of a brick chimney. If designed 
and built upon the same principles and by the same methods which 
have proved essential in other types of reinforced concrete construc- 
tion, they can be depended upon to give permanent satisfaction. 

Reports? on a large number of chimneys have shown that con- 
crete is unaffected by the heat from an ordinary steam boiler plant. — 
The temperature in such chimneys seldom exceeds 700° Fahr., 
while 400° to 500° Fahr. is more usual. Experimental tests also 
indicate that concrete is not appreciably injured by temperatures of 
600° to 700° Fahr.° 

To provide for extremes, it is advisable, however, to build an 
independent inner shell of concrete or firebrick for at least a portion 
of the height. Concrete should not be used for a chimney in con- 
nection with special high temperature furnaces. 

Since concrete and steel have substantially the same coefficient 
of expansion * there is no danger of heat causing a separation of the 
reinforcement from the concrete. 

2A special investigation of reinforced concrete chimneys was made . by 
Sanford E. Thompson in 1907 for the Association of American Portland Cement 
Manufacturers. Many of the points here discussed are summarized from the 
report, which is printed as Bulletin No. 18 of the Association. — ? 


3 Tests of Metals, U.S. A. 
4 See Volume IT of this treatise.. 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 823 


The expansive effect of heat is a more serious question. Stresses 
are set up in the shell of any masonry chimney, because of the hot 
interior and cold exterior surfaces. A concrete chimney, however, 
has thinner walls, and the stress is therefore less than in one of brick 
or tile; the concrete chimney is also better reinforced. Provision 
for temperature stresses are discussed in the paragraphs on design, 
which follow. 

Construction.—A reinforced concrete chimney, because of its 
height and shape, is more difficult to construct than many other 
kinds of concrete construction, and it therefore should be handled 
by experienced builders. 

It is essential in chimney construction that the materials be very 
carefully selected. The sand, as well as the cement, should be tested 
by determining the actual tensile strength of mortar made from it. 
The stone preferably should be of the nature of a hard traprock, 
3-In. maximum size. Proportions 1 : 2:3 have been found to give 
good results. A dry mix should not be used, since insufficient water 
will produce a porous concrete which does not adhere to the steel. 
The consistency must be wet enough to quake and form a jelly-like 
mass when lightly rammed, so as to properly imbed and bond the 
reinforcement. No exterior plastering should be permitted, because 
it is likely to check and scale. The steel should be round or deformed 
bars of good quality. Bars with flat surfaces, like T-bars, are inferior 
because the flat surfaces give a poor bond and the angles make the 
placing of the concrete difficult. Deformed bars of small size, quite 
closely spaced, are especially good for the horizontal steel to dis- 
tribute the temperature stresses, and high-carbon steel of first-class 
quality also has advantages for the horizontal reinforcement. 

Example.—The design of a chimney built in Brooklyn, N. Y., in 
1907, is illustrated in Fig. 294, p. 824. 

Design of Reinforced Concrete Chimneys.—A reinforced con- 
crete chimney consists primarily of a concrete shell with vertical steel 
bars disbributed uniformly along the circumference of the chimney. 
The shell must be of proper thickness and the steel bars sufficient in 
size and number to withstand the stresses due to the weight of the 
chimney and to the action of the wind. A chimney of this type 
differs essentially from one of brick, in that the diameter at the base 
is so small as compared to the height that the chimney would over- 
turn under a heavy wind were it not for the vertical bars of steel 
which serve as anchors and hold it on the windward side. 


REINFORCED CONCRETE CHIMNEYS 


824 


Section C-D 





eee Aree a ee a a a ae peer ee oh ee ren ce adh) 


© Q $23/1NO IV 





ind) 
ict n bent ered Senne Deeion cea o8 Ua AT 
nO-19F pe aes 0-08) 2 20706 = 7) 10nCe 














Section AB 





(See p. 823.) 






S 
Fic. 294.—Design of Chimney of the Edison Electric Illuminating Co. 


INS 












--16 
-----28'-"---- 
Brooklyn, 


; 


> t= So a ee re = gee Oe .. -—— 


x 
S4o0g 4a, gf | S4I2G 424 001 la 72 910) GA ool alt 


beat OS: 710~06 n0-OL n070L 10701 nG TOL 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 825 


Wind, in blowing against a chimney, causes compression on the 
side opposite the wind and tension on the side against which the 
wind is acting. This compression is resisted by the concrete and 
steel on the leeward side, while the tension or pull is taken by the 
steel on the windward side. 

In addition to the vertical reinforcement, a reinforced concrete 
chimney should be provided with horizontal hoops of steel, the object 
‘of which is to stiffen the vertical steel, to distribute cracks in the con- 
crete due to a difference in temperature between the interior and 
exterior, and to resist the diagonal tension. 

In designing a reinforced concrete chimney, the problem, then, is 
primarily to determine at various horizontal sections the necessary 
thickness of the concrete shell and the required amount of vertical 
reinforcement, so that the allowable working stresses in the concrete 
and in the steel shall not be exceeded under the action of the forces to 
which the structure may be subjected. The complete analysis and 
development of the most useful formulas are given in the first part 
of this chapter. The final formulas to be used in design are repro- 
duced below. 

The problem of the determination of stresses due to the difference 
in temperature between the interior and the exterior of the shell 
involves many uncertainties. The heat tends to expand the inner 
surfaces, producing tension in the outside surface of the shell and 
compression in the interior surface. Although the distribution of 
the stress is not clearly known, the variation of the heat through the 
shell not being uniform, tentative computations indicate high stresses, 
so that it is a question whether vertical temperature cracks can be 
entirely prevented any more than they can be prevented in brick or 
tile chimneys. The function of the horizontal steel may therefore 
be to distribute these cracks and to resist the vertical shear or diagonal 
tension. This horizontal steel should be distributed, therefore, by 
using small diameter bars closely spaced rather than large bars 
spaced further apart. Because of the possibility of vertical tem- 
perature cracks, the concrete should never be relied upon to carry 
tension or vertical shear, and the amount of horizontal reinforcement 
to resist this may be obtained in a fashion similar to the determina- 
tion of vertical stirrups in a beam. The analysis for the shearing 
stresses is given on page 819, and the final formula is presented below 
together with suggestions for adapting the horizontal reinforcement to 
temperature stresses. 


826 REINFORCED CONCRETE CHIMNEYS 


The amount of vertical reinforcement, the thickness of the shell, 
and the percentage of horizontal reinforcement may be obtained 
from the following formulas. 


Let W = weight of the chimney above the section under considera- 
ton, 1b.) 
M = moment of the wind about that section, in.-lb.; 
f, = maximum tension in the steel, lb. per sq. ak 
f. = maximum compression in the concrete ineadived at the 
mean circumference), lb. per sq. in.; 


s 


n= E, = ratio of modulus of elasticity of steel to that of con- 


crete; 
D = mean diameter of shell (i.e., diameter of center of ring), in.; 
r = mean radius of shell, in.; 
t = total thickness of shell, in. ; 
A, = total area of vertical reinforcing bars in the section under 
consideration, sq. in.; 
k = ratio of distance of rauriat axis, from mean circumference 
on compression side, to the mean diameter D; 
2, Cp, Cr = constants for any given value of k, Tables 1 and 2, p. 
828 ; 
po = ratio of cross-sectional area of steel hoop to vertical sec- 
tional area of concrete; 
h, = height of chimney above section under consideration, ft.; 
F = effective wind pressure against chimney, lb. per sq. it.; 


Then 
Area of vertical bars 
BOE Wa2d) 3 
Ale rae Crf.D e e e e e e e e e e (9) 


Thickness of shell 


A, 
2W + (Crfs — Cefn) — a 


eee ge we A 


“aa Cif.D xD 





Ratio of steel hoops 


hiP 
Po = agrgey + 0.0025: 1 (11) 


Formulas (9), (10), and (11) correspond to Formulas (6), (7), 
and (8), pp. 818 to 821. 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 827 


In the formula for po, the first term gives the ratio of steel to 
resist vertical shear or diagonal tension, and the second term is an 
arbitrary ratio designed to distribute the temperature strains. To 
best distribute the temperature strains, a maximum spacing of 6 in. 
to 10 in. is recommended for the horizontal bars. 

In the formulas, the terms z, Cp and Cy are constants, the values 
of which are fixed for any given position of the neutral axis. By 
means of Tables 1 and 2 (p. 828), these constants may be easily 
and quickly determined, so that the solution of Formulas (1) and 
(2) is rendered quite simple after the diameter and height of the 
chimney have been selected and the bending moments due to the 
wind have been computed for the various sections considered. _ The 
thickness of shell must be assumed in Formula (1) in order to deter- 
mine the average diameter, D, and to compute the weight, W. A 
new computation may be made to correct this, if necessary. For 
economical distribution of concrete and steel, computation must be 
made for several sections in the height. It is not advisable to make 
the thickness of exterior shell less than 5 in. at any place, but the ~ 
number of steel rods may be gradually reduced toward the top. 

Summary of Essentials in Design and Construction—In the 
investigation ° referred to, the essential requirements are sum- 
marized as follows: | 

(1) Design the foundations according to the best engineering 
practice. 

(2) Compute the dimensions and reinforcement in the chimney 
with conservative units of stress, providing a factor of safety in the 
concrete of not less than 4 or 5. 

(3) Provide enough vertical steel to take all of the pull without 
exceeding 14 000 or at most 16 000 Ib. per sq. in. 

(4) Provide enough horizontal or circular steel to take all the 
vertical shear and to resist the tendency to expansion due to the 
interior heat. 

(5) Distribute the horizontal steel by numerous small rods, in 
preference to larger rods spaced farther apart. 

(6) Specially reinforce sections at which the thickness in the wall 
of the chimney is changed, or which are liable to marked changes of 
temperature. 

(7) Select first-class materials and thoroughly test them before 
and during the progress of the work. 

> See footnote, p. 822. 


828 


Table 1. Values of Constants Cp, Cr, 

Neutral Axis, (i.e., for various values of k) 
(5), (6) and (7), pages 818 and 819, and (9), (10), and 
stance of neutral axis from mean circumference 
diameter D. Value of k to suit the condition 


For use with equations 
(11), page 826. & is ratio of di 
on compression side to the mean 


REINFORCED CONCRETE CHIMNEYS 


of the problem is obtained from Table 2, below. 





i) 
is) <> 
On 
oO 


100 
.150 
200 
250 
300 
390 
400 
450 
500 
.590 
.600 


Table 2. Location of Neutral Axis for Vario 








0 
0 


DONNER err e 


Cp 





.600 
852 
.049 
.218 
.310 
.510 
.640 
765 
.884 
.000 
113 
224 








Cr Z 
3.008 ' 0.490 
2.887 0.480 
Dat h a 0.469 
2.661 0.459 
2.551 . 0.448 
2.442 0.438 
2.333 0.427 
2.224 0.416 
2.113 0.404 
2.000 0.393 
1.884 0.381 
1.765 0.369 





J 


0.760 
0.766 
0.771 
0.776 
O.779 
0.781 
0.783 
0.784 
0.785 
0.786 
0.785 
0.784 


Stress, f,, Tensile Stress, /s and Ratio of Moduli, n. (See p. 827.) 


Maximum 
Tensile Stress 
in Steel, fs 





ee 


k Ratio of Depth of Neutral Axis to Depth of Steel Below Most 
Compressed Surface of Beam 


m = 10 


eee as 


Maximum compressive 
stress in concrete, fc 


 n=i12 





————_—_ 


Piet) ie) 


z and j for Different Positions of the 


a 


us Combinations of Compressive 





Maximum compressive Maximum compressive 


stress in concrete, fc 




















stress in concrete, fc 





300 | 400 | 500 | 600 | 700 | 300 | 400 | 500 600 | 700 





-310| .375| .428] .474| .512) .360). 
285] .348] .400| .444|.483) 334]. 
264| .324| .375] .418] .456) .310]. 
246] .304| . 353] .395] .433) . 290). 
231] .285| .334| .375} .412| . 272). 
217| .270| .316| .356} .392) .257/. 
-204| .255| .300| .340] .375| .243). 
198] . 242] . 286] .324|. 360) . 231). 
184] .231| .272| .310| .344) .220). 
175] .220| . 261] .298] .330} .210). 
166] .210] . 250) . 285} .318) . 200). 
160] . 201] . 240) . 275] .306) .192). 
152] .194| .231| .264| . 296) . 184). 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 829 


(8) Mix the concrete thoroughly and provide enough water to 
produce a quaking concrete. 

(9) Bond the layers of concrete together. 

(10) Place the steel accurately. 

(11) Place the concrete around the steel carefully, ramming itso 
thoroughly that it will slush against the steel and adhere at every 
point. 

(12) Keep the forms rigid. 

The fulfillment of these requirements will increase the cost of the 
structure; but if the recommendations are followed, there should be 
no difficulty in erecting concrete chimneys which will give thorough 
satisfaction and will endure. 

In connection with reinforced concrete chimneys, the problems 
which arise are of two general kinds: 

(1) A problem in design, involving the determination of the neces- 
sary thickness of shell and required amount of reinforcement at the 
various sections of a chimney of given height and diameter. 

(2) A problem in the review or investigation of a chimney of 
given height and diameter having a certain thickness of shell and a 
given amount of reinforcement, to determine the stresses in the 
concrete and the steel under the action of certain forces. 

The application of the foregoing formulas to such problems and 
the use of the accompanying tables may best be illustrated by the 
following numerical examples. 


EXAMPLES OF CHIMNEY DESIGN 


Example 1.—Given a chimney with height above section considered, 110 ft.; 
mean diameter at section considered, 10 ft.; allowable pressure in concrete (f,), 
500 Ib. per sq. in.; allowable tension in steel (fs), 14 000 lb. per sq. in.; ratio of 
moduli n, 15; wind pressure (on normal plane) 50 lb. per sq. ft., weight of con- 
crete taken as 150 lb. per cu. ft. What is the necessary thickness of shell and 
amount of reinforcement at the given section? 

Solution.—As in all chimney designs, it is necessary here to make a trial 
assumption of the thickness of shell in order to estimate the weight. Suppose 
Wwe assume a 6-in. shell for the entire height above the section. Assuming that a 
wind pressure of 50 Ib. per sq. ft. on a normal plane corresponds to 8 of 50 Ib. 
or 30 lb. per sq. ft. on the projected diameter of a cylindrical surface, we have 
the bending moment due to the wind, 


M = [10.5 X 110 X 30] X +32 X 12 = 22 869 000 in.-lb. 
and the total weight of the chimney above the section, 
W = 3.1416 X 10X 0.5X 110 X 150 = 259 180 lb. 


830 REINFORCED CONCRETE: CHIMNEYS 


f. = 500, fs = 14000, and n = 15, Table 1 gives k = .349. 
For 


ce 
I 


349, Table 2 gives Cp = 1.687, Cr = 2 335, 2 eae, 
Substituting in equation (1), 


_ 8(22 869 000 — 259 180 427 120) 
i 2.335 X 14000 X 120 





= 19.6. 


=* 5 


Therefore, 19.6 sq. in. of steel are required. 
If 3-in. round rods are selected, 45 of them will be required. 
Substituting in equation (2), we have 





19. 
2> 259 180 + [(2.335 X 14.000) — (1.637 X 500 X 15)l5 77a 





1.637 X 500 X 120 


19.6 
3.1416 X 120 





= 6.6 in. 


Therefore a 6.6-in. shell would be used. 

In general, the values of A, and ¢ as thus obtained should be readjusted by 
computing W on the basis of the computed thickness of shell. In the case at 
hand, however, the original assumption of a 6-in. thickness corresponds, for all 
practical purposes, to the computed thickness of 6.6 in., so that recomputation 
is, in this case, unnecessary. If the walls of the chimney taper in thickness, the 
value of W must be altered accordingly.® 

When the required thickness of shell and amount of vertical reinforcement 


have been determined, there remains the question of the necessary horizontal or ‘ 
circular reinforcement. Substituting in Formula (3) for fs, say, 14.000 lb., we © 


have 
110 X 30 


Po = 18.8 14000 X 6.6 





+ 0.0025 = 0.0044. 


Area of steel, As = 6.6 X 12 X 0.0044 = 0.35 sq. in. Thus 3-in. round rods — 


should be spaced 6{ in. on centers. 
In a similar manner, any other section of the chimney may be proportioned. 


REVIEW OF A CHIMNEY 


Example 2.—Given a chimney with height above section considered, 90 ft.3 @ 


, 


mean diameter at section considered, 8 ft.; thickness of shell at section con- — 


sidered, 6 in.; vertical steel at section considered, sixty 3-in. round rods; wind 
pressure (on normal plane, 50 lb. per sq. ft.); weight of concrete taken as 150 lb. 
per sq. ft.; ratio of moduli, n, 15. 


the section under consideration? 


6 In relatively high chimneys steel cannot be stressed to 14 000 lbs. per sq. in. (see p. 821.) 


a ee a ee 


q 
4 


What are the maximum stresses in the concrete and in the vertical steel at ; 


DESIGN OF REINFORCED CONCRETE CHIMNEYS 831 


Solution.—A problem of this kind must necessarily be solved by a method 
of successive trials, since the position of the neutral axis is not-known. The 
location of the neutral axis is determined by the values of f,, fs and n, two of 
which, in this case, are unknown. The method of procedure, therefore, is to 
assume outright a trial position of the neutral axis, select the constants accord- 
ingly, substitute in equations (1) and (2) and solve them for f, and foe 

Then see if the position of the neutral axis, as fixed by these values of Ts ana Ts 
and the given n, is the same as the position assumed at the start. If the two 
positions agree, then f, and f,.as found are the actual stresses; if not, a new 
position of the neutral axis must be assumed, new constants selected, and new 
values of f; and f, computed from equations (1) and (2). Thus a series of trials 
must be made until the location of the neutral axis as assumed is consistent with 
the computed values of f, and f, together with the given n. 

In this problem, assuming 30 Ib. pressure on the projected area, we have the 
bending moment due to the wind, 


90 
M = [8.5 X 90 X 30} x ms < 12 = 12 393 000 in. lb. 


and the total weight of the chimney above the section, 
W = 3.1416 X 8 X 0.5 X 90 X 150 = 169 646 lb. 
A, =60 X .3068 = 18.41 sq. in. 


Now suppose we assume the neutral axis at, say, k = .400 
For k = .400, table on p. 828, gives Cp = 1.765, Cr = 2.224, 2 = .416. 
Substituting in equation (1) we have 

8(12 393 000 - 169 646 x .416 x 96) 


18.41 = 
2.224 X fs X 96 





whence f, = 11 400. 
Substituting in equation (2) we have, | 
18.41 


3.1416 ee lord 
1765 f,0 96 3.1416 X 96 


2 X 169 646 + (2.224 x 11 400 — 1.765.f, 15) 








whence f, = 416. 


Now fs = 11 400, f, = 416, and r = 15 gives k = .354 which does not cor- 
respond with our original assumption of z = .400. Evidently the true & must 
lie somewhere between the assumed and determined values, hence if we now 
assume, say, & = .375 and recompute, we obtain fs; = 11000 and f, = 435, 
the values of which together with n = 15 gives k = .371 which checks fairly 
well with the assumption of k = .375. For all practical purposes we may 
therefore say that the maximum stress in the steel is 11 000 Ib. per square inch, 
while the maximum stress in the concrete is 435 lb. per square inch. The results 
indicate that both the thickness of shell and the amount of steel are greater than 
are necessary for safe stresses. 


CHAPTER XXI 
RETAINING WALLS 


For walls designed to resist the pressure of earth or water, con- 
crete is generally superseding other classes of masonry. In most 
localities, its cost is less than that of rubble masonry. Its adaptabil- 
ity for thin walls and for certain classes of face work often makes it 
a suitable substitute, in complicated designs, for first-class masonry, 
with a consequent large saving in cost. In combination with steel, 
its possibilities for special designs are almost unlimited. 

Water-tightness, often an essential element for this class of struc- 
tures, receives general treatment in Vol. II. Portland cement con- 
crete may be made water-tight more readily than stone masonry 
laid in mortar of similar proportions to the cement and sand in the 
concrete, since large voids or stone pockets in the concrete are more 
easily prevented than the “ rat-holes ”’ so frequently found in the © 
bedding of stones in mortar. Moreover, with careful selection of — 
aggregates and skill in laying, thinner, impermeable walls may be 
built of concrete—strengthened with steel reinforcement—than is 
possible with stone masonry. 

Reinforced concrete retaining walls cannot be designed by “rule 
of thumb,”’ and therefore a careful consideration of the forces acting — 
on them, and of the stresses in the concrete, is presented in this | 
chapter. Since.the earth pressure is the controlling factor, it will be 
necessary to introduce a practical discussion of this before taking up 
the details of the design and examples of the two principal types. 

Relative Economy of Plain and Reinforced Concrete Retaining — 
Walls.—Retaining walls to support the pressure of earth may be ~ 
designed: 


(1) of gravity section with plain concrete or stone masonry; 
(2) of thin reinforced concrete section with spreading base 
or footing. 


Reinforced concrete retaining walls are almost always more 
economical than gravity sections of either plain concrete or masonry. 
Sa. 


WEIGHT OF EARTH 833 


In the gravity section, the materials cannot be fully utilized because 
the section must be made heavy enough to prevent overturning by 
its own weight. Counterforts or buttresses are of comparatively 
little advantage because, in stone masonry, or even in plain concrete, 
‘the wall is likely to break away from them. In reinforced concrete 
retaining walls, on the other hand, a part of the sustained material 
can be used to prevent overturning, and the section need only be 
made strong enough to withstand the moments and shears due to 
the earth pressure. Since the wall is lighter, exerts less pressure on 
the soil, and may be made, if necessary, with a very broad base, the 
special foundations or piling that are often necessary for a gravity 
wall may frequently be avoided. Reinforced concrete, properly 
designed, can be depended upon as absolutely reliable. 

The economy of a reinforced concrete wall, over a wall of gravity 
section of either stone masonry or plain concrete, is obvious because 
of the saving in material. The cost of forms is practically the same 
for reinforced designs as for gravity section. 

Mr. J. I. Oberlander reports! that twenty-three bids submitted 
on alternate designs of gravity and T-shaped reinforced concrete 
sections showed the average total cost per linear foot for the gravity 
section to be about one-third greater than that for the reinforced 
section. The unit price for the concrete in the reinforced section, 
however, was about 20 per cent greater than that for the gravity 
section. | | 


WEIGHT OF EARTH 


In the calculation of retaining walls, and many other structures, 
the weight of earth in place is a prime factor. The weights of dry 
material, based upon experiments by the authors, are represented in 
the following table. Most of the figures for. weights of earth give 
the weights per cubic foot after excavation in a loose or a compacted 
condition. In the authors’ experiments, the excavation was meas- 
ured, so that the weights represent the material in place. As fills 
will eventually assume much the same characteristics as earth in 
original excavation, the figures may be employed for either natural 
earth or filled material. The weight of earth containing water varies 
with the character of the material and with the conditions. Gravel 
containing ordinary moisture weighs about 2 per cent more than 


1 Engineering and Contracting, May 19, 1915, p. 457. 


834 RETAINING WALLS 


dry gravel; and sand may weigh from 3 to 10 per cent more,” depend- 
ing upon its fineness, since the finest sands absorb the moist water. 
Wet muck weighs about 75 lb. per cu. ft: These percentages assume 
that the bank is provided with natural drainage; if the earth is 
literally filled with water which cannot run off its weight will be 
increased by a quantity of water nearly equal in volume to the 
voids in the material, which vary, with the character of the material, 
from 20 to 50 per cent of the bulk of the earth in the bank. 

Many of the values appear high, but they are the result of care- 


ful tests. 


Average Weight of Ordinary Earth Before Excavation — 


Lb. per 

Cu. Ft 

Cand oO eee 105 
CGravelina J) 110 igs 2 135 
Gravelly clay ..0si.1> 5-00 e) Se = ae 130 
TiO aT ce ed’s wid eomeael « Card nln a eal ee eee 90 
Hardpan «...-..+ 9+ ¢17 9 oar oiae 2 een 130 
Dry muck 7 uo... ers se AQ 


Drainage of Retaining Walls.—The drainage of retaining walls 
is a highly important matter, and lack of drainage may cause either 
complete or partial failure. A common plan is to lay a drain of tile 
or of broken stone along the back of the base, opening at the ends of 
the wall or discharging through weep-holes. 

The Delaware, Lackawanna & Western R.R.® builds 4-in. tiles 
through the wall at frequent intervals along the footing, with a right- 
angle elbow, turned up, on the inner side. A chimney of loose rubble, 
about 2} by 3 ft., runs from each weeper up to the top of the wall. 

The Rock Island Railroad,* in some track elevation work, laid a 
line of tile drain along the back of the wall and carried the water 
through the abutment by a weeper running to a storm sewer. 

Mr. Lindenthal’s high walls (see p. 873) were drained through 
weep-holes, placed every 10 ft., through each wall. From each weeper 
one 4 by 4-ft. dry rubble chimney was built up back of the wall to 
the surface. 

2 This larger weight applies to moist sand in place. Loose, moist sand on the 
other hand, as stated in Vol. II, weighs less than dry sand. ; 


3 Engineering Record, Jan. 3, 1914, p. 29. 
4 Engineering News, April 8, 1915, p. 670. 


EARTH PRESSURE 835 


Another good method of drainage consists of a layer of loose 
stone over the entire back, with weep-holes placed at intervals. 

Frost.—The depth of foundation must be sufficient to prevent 
heaving of the material in front of the wall, and to protect it from 
frost. A depth of 3 ft. may be given as a minimum, while 4 or 5 ft. 
is necessary in temperate or very cold climates. 

Even with the base safely below frost level, special precautions 
are sometimes necessary to prevent heaving by frost-grip on the side 
of the wall or abutment. Such a case, cited by Edward H. Rigby,5 
was encountered in China, where frost gripped the side of bridge 
abutments to a depth of 5 ft. and lifted them, railroad, girder bridges, 
and abutments, clear off the pile foundations. Piers and abutments 
with sloping faces were lifted as much as those with vertical faces. 
Computations showed that the average lifting power of the frost 
was 1 000 lb. per sq. ft. of exposed surface, and that the remedy was 
to design the piers and bridges to overcome this force by dead weight. 


EARTH PRESSURE 


The principal force governing the dimensions of a retaining wall 
is the earth pressure. The magnitude of the earth pressure depends 
upon the character of the soil, its moisture content, the slope of the 
earth back of the wall, and also the inclination of the back surface of 
the wall. 

Rankine’s Theory.—The magnitude of the earth pressure is 
affected by so many variables, that it is incapable of being solved by 
an exact method applying to all conditions. The best theory thus 
far advanced is Rankine’s, based on the assumption that the earth 
is composed of granular particles without cohesion, held together only 
by friction developed between them. 

Direction of Earth Pressure.—The unit earth pressures—also 
the total earth pressure—are assumed to act in a direction parallel 
to the slope of the ground (for level ground or aestheley sloping up). 
This is shown in Fig. 295, p. 836. 

Distribution of Earth Pressure.—The unit earth pressure is 
assumed to increase proportionally with the depth below the ground 
level. The variation of the pressure may be represented by a tri- 
angle, as shown in Fig. 295. The unit pressure is zero at the ground 
level and a maximum at the bottom of the wall. 


5 Engineering News, March 5, 1908, p. 260. 


836 RETAINING WALLS 


Let P = resultant earth pressure in lb. on a vertical surface for 
a length of wall equal to one ft.; 

h = total height of wall, ft. ; 
h; = depth below top of wall of any point, ft.; 

w = weight of earth, lb. per cu. for | 

6 = angle of inclination of earth behind the wall; 

® = angle of internal friction of the earth; 
C, = constant depending upon 6 and ¢. (See table on p. 837.) 





——---—--—-},—-—----—- 
4. Mioep Be 


Cp wh 
les 
ecVerres wey We 





Value of Cp 1s different in the two cases 


Frc. 295.—Distribution of Earth Pressure. (See p. 835.) 


Unit Pressures on Vertical Surface.— 
Unit Pressure at Any Depth, hz, below Ground Level, 


_| Lape Shea 
cos 6 — Vcos? 6 — cos? ¢ 














x= w cose i 
: cos 8 Vcos? 6 — cos? () 
calling 
Ciel ans 5 008 5 — Vos? 6 — cos? ¢ 
: cos 6 + V cos? 6 — cos? ¢ 
a = Cytthe! » + +o ee ee 
Maximum Unit Pressure at Bottom, 
p = Cwh...> ig ee ee : 3G 


Values of C, are given in table, p. 837. 

Special Cases of Earth Pressure.—The above formula is general. 
Below are given formulas for special cases. 

When ground is level, 6 = 9, and the constant C, changes to 
1 —sin¢ 


oar 1+ sin ¢ 


EARTH PRESSURE 837 


When ground slopes at angle equal to angle of repose (6 = ¢), 

the constant becomes 
Cz = cos 6. 

Total Pressure.—Since the pressure is represented by a triangle, 
the total pressure is obtained by multiplying the maximum unit 
pressure at the bottom by one-half of the height. The point of 
application of the pressure is at one-third of the height from the 
bottom. Thus, 

Total Pressure, 

Tei Ri, ae eo LS te 1, (A) 


Values of C, may be taken from table below. 


Values of Constant Cp. 
cos 6 — V cos? 6 — cos? } 
cos 6 + Vcos? 6 — cos? b 








Cp = cos 6 





Values of Constant Cp 





Slope with Horizontal 









































Angle ; 
f 
Pato | 1to1s | 1to2 | 1t023| 1to3 | 1to4 | Level | 
Friction ! 
? Corresponding Angle of Slope 6 
| 
fo ego a0 e206, 40’ 121° 50’ | 18° 30% | 14° 0 0 p 
55° 0.18 0.13 0.12 0.11 0.11 0.10 0.10 | 0.57 
50° 0.29 0.18 0.16 0.15 0.14 0.14 0.13 | 0.64 
Tee hee... 9 0.26 0.22 0.20 0.19 0.18 OPDD. WOT 
ae 0.36 0.29 0.26 0). 24 0.23 O22 O00 
SOOM shane. 0.58 0.38 0.33 0.31 0.29 0727 |) 0782 
SoU) sie Meet pit 6 1b ire, Svs xi ay 0.54 0.44 0.40 0.37 0733 |, 0°87 
EET VAS EE Ss Per 0.60 0.52 0.46 0.40 | 0.91 
RMIT Aee Ma is ras to) ec lckia wey os 0.72 0.58 0.49 | 0.94 


























Nortr.—If the angle of internal friction of the earth is unknown, the following 
average values may be used: Coal, shingle and broken stone, 50°; earth, 35°; 
clay, 30°; sand dry, 30°; sand moist, 35°; sand wet, 20°. 


Surcharge.— When the earth behind the wall is loaded in any way, 
as for example, when a highway or a railroad track runs along the 


838 RETAINING WALLS 


wall, or when the embankment is used for the storage of material, 
this loading causes additional pressure on the wall, which should be 
provided for by replacing the 
weight of the load by an equiva- 
lent surcharge of earth. The 
height of this surcharge, hi, is the 
load per square foot divided by 
the weight of a cubic foot of earth. 
Thus, a load of 500 Ib. per sq. ft. 
is equivalent to a surcharge of 5 
ft., if the earth weighs 100 Ib. per 
eu. It, | 

The earth pressure on a re- 
taining wall with surcharge is 
Fia. 296.—Earth Pressure with Sur- represented by a triangle with the 

chatge, | (See P8338.) zero point at the top of surcharge. 

The maximum pressure at the 

bottom is p = C,w(h + *1), where hi is the distance from level 
eround to top of surcharge. (See Fig. 296, p, 838.) 

The pressure resisted by the wall is represented by a trapezoid, 
the top of which is C,wh; and the bottom C,w(h + hi). The total 
pressure equals the area of the trapezoid. 

In addition to notation on p. 836, 

Let h, = distance from level of ground to top of surcharge. 

Total Pressure with Surcharge, 





ahr 


si ma te eres Bk 


ae Chath) 


P-= iC w(h + 2h 
Height of Point of Application of Pressure above Bottom, 


he Bhi h 
= FL, 3 oe Ree tay 





x 


Wall with Inclined Back.—The earth pressure, #, on an inclined 
plane ab (Fig. 297) is the resultant of P, the horizontal pressure on 
the vertical plane ac, and W, the weight of the prism of earth abc, 
and acts at one-third the height from the bottom. 

Resistance of Wall to Sliding—The horizontal component o 
the earth pressure causes the tendency of the wall to slide on its base. 
This force is opposed mainly by the friction between the base of the 
wall and the earth, which is equal to the vertical pressure multiplied 


DESIGN OF RETAINING WALLS OF GRAVITY SECTION 839 


by the tangent of the angle of friction between concrete and earth. 
Let . 

F = total frictional resistance of the base; 
Wi + W2 = weight of concrete and earth above the base; 

N = vertical component of earth pressure; 


y = angle of friction between earth and concrete, 
Then 
F = (Wi+ W2+ N) tan y. Woke Site) LeWi eat) 26) Ne ts ah) 


If the friction is larger than the horizontal component of earth 
pressure multiplied by a factor of safety, the wall is secure against 
sliding. This rule may also be expressed as 
follows:® If the angle which the resultant pres- 
sure makes with the vertical is smaller than 
the angle of friction, the wall is safe against 
sliding. . 

The resistance to sliding is increased by the 
passive pressure of the earth in front of the wall, 
and particularly in front of the toe. Also, the 
cohesion of the earth back of the wall assists the 
resistance to some extent. 

In many cases, the resistance to sliding is in- 16: 297-— Harth 

‘ ees Pressure on Inclined 
creased by a vertical downward projection of the Back of Wall. 
base or a key, which may be placed in the middle (See p. 838.) 
of the base or at its heel end. 

Passive Earth Pressure.—The passive pressure of a mass of earth, 
that is, its resistance to movement or compression, is many times as 
great as the active pressure. Because of the possibility of shrinkage 
of earth fill in front of the wall, away from the wall, the passive 
pressure of the earth in front of the wall cannot be counted on to 
counteract active pressure due to the earth behind the wall 





DESIGN OF RETAINING WALLS OF GRAVITY SECTION 


The thickness of gravity retaining walls is frequently determined 
by rule-of-thumb, but this is an unsatisfactory procedure. On 
work of any importance, much more economical results are obtained 
by special designs, governed by the character of the foundation soil and 
the earthbacking. Partial failures—tipping forward, cracking, and 

6 The horizontal component equals (Wi + W2-+ N) multiplied by the tangent 


of the angle of the resultant with the vertical. If this angle is smaller than y, 
its tangent will be smaller; therefore, the horizontal force is smaller than friction. 


840 RETAINING WALLS 


sinking—are prevalent among retaining walls. In one case, a heavy 
gravity wall failed under the weight and impact of the backfilling 
dropped through a distance of 30 or AO ft. from a large drag line scraper 
bucket. Where the foundation is so poor that such action is possible. 
the line of pressure should pass through the base, well within its 
middle third. Uneven settlement is then less likely to take place; 
and in any event the line of pressure has more opportunity to shift, 
without causing either uplift or excessive pressure on the base, than 
if the line passed through the forward edge of the middle third.’ 

The methods of determining the line of pressure and of computing 
pressure on foundation are similar to those discussed in connection 
with reinforced walls. 

If empirical rules are to be used, one that is easily remembered is 
to make the base three-eighths of the height, for walls without sur- 
charge. Another is to make the base at least as thick as would be 
necessary if the wall were to be subjected to water pressure under a 
head two-thirds the height of the wall. A table of empirical values, 
adopted by Mr. Trautwine for thickness of base of masonry walls to 
resist earth pressure, is given below. 


Thickness of Retaining Walls of Gravity Section with Earth Surcharge 
By Joun C. TRAUTWINE. (See p. 840.) 




















Ratio of Thickness of Ratio of Thickness of 
Height of Earth to | Base as Ratio to | Height of Earth to | Ease as Ratio to 
Height of Wall Height of Wall Height of Wall Height of Wall 
1 0.35 2 0.58 
a | 0.42 275 0.60 
1,2 0.46 3 0.62 
1.3 0.49 4 0.63 
1.4 0.51 6 0.64 
1.5 0.52 9 0.65 
eae) 0.54 14 0.66 
hare 0.55 25 
1.8 0.56 or more 0.68 





ee ee 
Walls designed by these empirical methods are unsafe under 
unusual pressures, such as quicksands, and detailed analyses must 
7 Certain failures of this type are discussed by Charles K. Mohler in Engineer- 


ing News, Oct. 18, 1910, p. 384. 
8 Suggested by Engineering News, Sept. 26, 1912, p. 593. 


DESIGN OF REINFORCED CONCRETE RETAINING WALLS 841 


be made. On the other hand, the designs obtained by empirical 
methods are in many cases unnecessarily conservative. 

The height of the wall is assumed to be measured above the sur- 
face of the firm ground in front of it. 

The batter of the face of a retaining wall is customarily limited 
to 13 inches to the foot, and the back is usually vertical. This fixes 
the width on top. 

The values in the table may be employed for long walls of con- 
crete with no reinforcement. In many cases, because of the mono- 
lithic character of concrete, a ratio of thickness to height from 10 to 
20 per cent less may be adopted with safety, if the character of the 
filling back of the wall precludes excessive pressure, and if the base 
is slightly spread. For more accurate determinations of gravity 
sections, the principles which follow, relating to reinforced designs, 
are applicable. When two walls enclose a narrow fill, they may be 
tied together by rods, as discussed on p. 872, and thinner sections 
used. Similarly, the ordinary single wall may be anchored to the 
ground behind it. 


DESIGN OF REINFORCED CONCRETE RETAINING WALLS 


General Principles.—A reinforced concrete retaining wall may 
fail as a whole, if it settles excessively or slides on its base. Its 
failure may also be caused by the failure of its component parts. 
For instance, in a cantilever wall, the vertical slab may break away 
from the foundation, or the toe cantilever may crack or shear off. 

To prevent the failure of the wall as a whole, the base must be 
properly proportioned. In a properly proportioned retaining wall, 
(1) the pressures on the foundation must be within the allowable 
working limits, and must be properly distributed over the base; and 
(2) there must be sufficient resistance to sliding of the wall. 

To prevent failure of the component parts, each part must be 
designed according to the accepted rules, to resist the stresses to 
which it is subjected. Also, the parts must be properly connected: 

Earth Pressure to be Used in Design.—The base of a retaining 
wall is usually several feet below the level of the lower ground, in 
order to protect it from frost and to reach solid ground. There is, 
therefore, a certain amount of earth in front of the wall. In fixing 
the amount of earth pressure to which a wall is subjected, the question 
arises whether it would not be permissible to utilize the passive 


842 RETAINING WALLS 


pressure of the earth in front of the wall to resist a part of the earth 
pressure. 

This is not permissible. The earth in front of the wall is usually 
a fill, which, until it becomes compacted, has very little passive 
resistance. Furthermore, the fill may shrink away from the wall. 
Finally, the fill may not always be in place at the time the earth 
pressure acts. A good rule is to design the wall for the amount of 
earth pressure acting above the top of the slab at the end of the 
base. (See Fig. 298 below.) 










Total Earth 


Pressure : 
i y Pressure 
/ 


on upright 









Pressure 
on upright 


‘ 


Total } 
ressure\ 


¢ 


{ 

' 

{ \ 

H ; 

| ’ 

| ' 

- 

\ wm : 
{ U t 
1 

| Resistance | Resistance / 
| of this earth of this earth 
unreliable y | unreliable \ 
| NTN | FARTING : 
| ’ \ 
! ; 5 { 
\ nie: ‘ ! 


@ Level Ground Reinforcement not shown. _(b) Sloped Ground 
Fic. 298.—Earth Pressure on Reinforced Concrete Retaining Walls. (See p. 842.) 


For walls supporting earth with a sloped surface, the height of the 
wall to be used in the formulas for earth pressure is the distance from 
the top of the slab at the end of the base to the top surface, measured 
on a vertical line. This distance 1s marked h; in Fig. 298. The 
width of base should be designed for this earth pressure. The amount 
of earth pressure to be used in designing various parts of the wall is 
discussed under the proper headings. 

Factor of Safety for Retaining Walls.—In’ all self-contained 
retaining walls, the earth pressure is resisted by the vertical loads 
acting on the retaining wall. It is a common characteristic of rein- 
forced concrete retaining walls that they utilize not only their dead 
load, but also the weight of earth resting on their base, to counteract 


DESIGN OF REINFORCED CONCRETE RETAINING WALLS 843 


the earth pressure. The amount of available weight of earth depends 
upon the width of the base back of the wall. The distribution of the 
pressure on the foundation depends upon the length of the toe, 
1.e., the part of the base in front of the wall. For any accepted 
dimensions of the base, the available vertical loads are fixed. 

To get the pressure on the foundation, it is necessary to decide 
upon the proper assumption as to the earth pressure. The combina- 
tion of the expected earth pressure, found from Formula (4), 
p. 837, with the vertical loads, would give the pressure on the founda- 
tion for working conditions only. This, however, does not solve the 
problem completely. 

The retaining wall should be investigated for accidental earth 
pressure, the magnitude of which is equal to the expected earth 
pressure multiplied by a selected factor of safety. The object is 
to make the wall stable not only for working conditions, but also to 
insure a definite factor of safety, just as in other structures. Ordi- 
narily, structures are subjected to one kind of loading. After the 
stresses at working loads have been computed, the stresses at increased 
loads may be computed by simple multiplication. If proper working 
stresses are adopted, the problem of the factor of safety is automatic- 
ally solved (see p. 125). Retaining walls, on the other hand, are 
subjected to two kinds of loading, largely independent of each other. 
The pressure on the foundation is a resultant of the earth pressure 
and the weight of the wall and of the earth above the base. The 
earth pressure may increase without necessarily increasing the 
vertical loads. For twice the earth pressure, the pressure on the 
foundation will not be doubled, but instead its distribution will be 
altogether changed. The point of application of the resultant on the 
foundation for increased earth pressure will be moved towards the 
toe, as is evident from Fig. 303, p. 857. The difference between the 
maximum and minimum pressures will be larger. 

The authors recommend that all walls shculd have at least a 
factor of safety of 2, as far as pressure on foundation is concerned. 

Allowable Unit Pressures on Foundation and Allowable Distri- 
bution.—The proper selection of a base requires some Judgment, 
as it is impossible to make rules applicable to all conditions. In no 
case, of course, should the pressures on the foundation for working 
conditions exceed safe bearing values on soil, as given on p. 471. If 
it is impossible to support the wall on soil within a reasonable depth 
below the surface, piles may have to be used, 


844 RETAINING WALLS 


For yielding soil, the pressure on foundation for working condi- 
tions should be distributed as uniformly as possible. Therefore, for 
expected earth pressure, the resultant — 
should be well within the middle third, 
but for earth pressure multiplied by the 
factor r = 2, the resultant should strike 
the base at the outside edge of the middle 


K L\ RaVEAW : 
Rath 7 a 
be <8 





2 third. The factor r is in this case 
! equal to the factor of safety. For un- 
TENTS yielding soils, such as hard pan or rock, 


the resultant for working loads may be 
allowed to intersect the base at the 
edge of middle third. In this case factor 
,y = 1. For earth pressure times factor 
of safety of two the resultant would be 
outside the middle third and there would 
} be theoretically, tension (or rather uplift) 
‘Pressure on foundation gn the outer edge, but the pressures at the 
Fic. 299.—Proportions of T- ¢o¢ would still remain within safe limits 
seen Segre BAY and therefore the wall would not fail. 

For intermediate conditions of the supporting soil, intermediate 
values of r (between 1 and 2) should be accepted for which the result- 
ant is allowed to reach the edge of middle third. 

The ratio r must not be confused with the factor of safety. 

Formulas for Width of Base——After the magnitude of the earth 
pressure is decided upon and the ratio, 7, is selected, the approximate 
width of base may be computed from following formula: 


“Let b = width of base in feet, ft.; . 
h.= height of wall from bottom to top of ground, for level 
eround; distance from bottom to top of surcharge, for 
wall with surcharge, feet; 

h, = distance from base to top of ground measured on a 
vertical erected at outside edge of base, for wall with 
sloped ground, feet; 

distance of point of application of forces R and N from 
center of base, ft. 
m = ratio of length of toe (measured to outside face of wall) 
to width of base. Fig. 299); : 
selected ratio of earth pressure, for which the resultant 
‘s allowed to reach outside edge of middle third, to 
expected earth pressure. 


© 
II 


3 
II 


DESIGN OF REINFORCED CONCRETE RETAINING WALLS 845 


Cy = constant for determining earth pressure, p. 837; 
Cy = constant for determining width of base, p. 845. 
Width of Base (approx.), when resultant strikes edge of middle 
third, 





b= 0.944) h for | . 4 ae 
Gre EE hs or level ground, (8) 





os rC, cos 6 
b = 0.94 ee 1 Si a for sloped ground, . . (9) 





or 


b = Ch for level and sloped ground, 


where C; is a constant from the table below, depending upon the 
ratio of toe to base, m, the constant of earth pressure, Cy, from 
_table on p. 837, and the selected ratio r, for which the resultant 

should strike the edge of middle third. For sloped walls, the value 
of C, should be taken from the table, that value being used which 
corresponds to the value of C, Pte icd by the cosine of the angle 
of slope. 


VALUES OF CONSTANTS C;, 


Teen od if ae Sa 
Gk Semana 





Values of rCp for Level Ground and rCp cos 6 for Sloped Ground 
























































Value 

of 

m | | | 

0.05)0.10/0.20/0.30/0.40|0.50\0.60:'0.70/0.80/0.90/ 1.0] 1.2}1.4]1.6]1.8]/ 2.0 
0 0.21)0.30/0.42/0.52'0.59/0.67)0.72/0.79/0.84/0.89)0.94/1.03!1.11/1.18]1.2611.32 
0.05 0. 20/0. 29|0.40/0.49/0.57/0.64|0.69/0.75/0.80!0.85/0.90'0.99/1.06/1.13/1.20/1.27 
0.10 0.19|0.28)0.39}0.48/0.55/0.62/0.67/0.73/0.77/0.82|0.87'0.96]1.02/1.09]1.16)1.22 
0.15 0.19|0.27/0.38/0.47|/0.53/0.60|0.65/0.71/0.75)0.80,0.85/0.93/1.00)/1.07/1.13/1.19 
0.20 0.18!0.27/0.37/0.46/0.52/0.59|0.64/0.70/0.74/0.79/0.83/0.91/0.98/1.05}1.11]1.17 
0.25 0.18/0.26/0.37|0.45/0. 52/0. 58)0.63/0.69/0.73/0.78)0.82/0.90/0.97/1.03)1.10/1.16 
0.30 0.18'0.26/0.37|0.45)0.51/0. 58/0.63/0.68/0.72/0.77/0.8110.90/0.96]1._03}1.09}1.15 
Ueao 0.1810. 26!0.3710.45/0.51/0.58)/0.63/0 68/0.72/0.77/0.81/0.89|0.96/1.02}1.09]}1.15 
0.40 0 18/0. 2610.37 0.45/0.52 ee ai 0.73)0.78'0.82/0,90/0.96/1.93/1.10}1.15 
| | | 











If it is desirable to get uniform pressure on the foundation, the 
resultant should strike the base in the middle. For such a condition, 
the formula for width of base becomes: 


S46 RETAINING WALLS 


Width of Base (approx.), when resultant strikes in the middle 


C 
ae Ram bY nat 
b 0.55) aryl for level oTOuNid) <a. ene : (10) 
Cy cos 6 
= Pred ace Poe panes 
b = (0155 mig) re a sloped ground; ". .. . (Ey 


where r, Cp, and m are same as explained on page 845. 

Pressure on Foundation.—Having determined the width of base, 
the pressure on foundation must be checked either by the approx- 
imate formulas (12) and (13) or the exact formulas (14) and (15). 
The approximate formulas look more complicated than the exact 
formulas, but they can be used directly, while the values NV and e 
must be found before the exact formulas can be used. In important 
work exact formulas should be used. | 

Approximate Pressure on Foundation.—The pressure upon earth 
may be computed approximately from following formulas. 


Let, in addition to notation on p. 844, 
w = unit weight of earth, lb. per cu. ft. ; 
p, and 2 = pressure at the toe and heel, respectively, lb. per sq. ft. 
Then 
Maximum Unit Pressure at Toe, Level Ground (approx.), 


M1 = wh} 1.10 — m)(1 — 3m) + 0.9720,(7) |. mend Ge). 


M inimum Unit Pressure at Heel, Level Ground (approx.) 


po = wh [1.10 — m)(1+ 3m) — 0.972C, () | ee. 


The formulas above give fairly accurate results for level ground. 

Exact Determination of Pressure on Foundation.—The pressure 
on foundation is best determined by semi-graphical method. First, 
the dimensions of the wall are assumed and drawn to scale. The 
weights of the component parts are computed and shown as applied 
at their respective centers of gravity. Next, the earth pressure is 
computed and shown as applying at one-third the height above the 
base. Finally, the forces are combined graphically by the well- | 
known method. The resultant is produced to the bottom of the 
base, and is then resolved into vertical and horizontal components. 
The vertical component produces pressure on the foundation, while 
the horizontal component produces a tendency to slide. 


DESIGN OF REINFORCED CONCRETE RETAINING WALLS 847 


After the point of application of the resultant has been found, 
the unit pressures at the toe and at the heel may be found from Ants 
following formulas. (See Fig. 299, p. 844.) 

Let e = distance of point of eenhcetion of forces R and N from 

center of base, {t.; 
N = normal component of the resultant, R, lb. 

Then 

Pressure at toe, 


p= ae = .) JOP HCE BOE b. aad tw werpe(oldd 
Pressure at heel, 
Do = sae — *) ibepereqaft.s tad. v7 tip) 


For graphical method see Fig. 303, p. 857. 

Derivation of Formulas (8) to (13).—In Fig. 298, the earth 
pressure is (Formula (4), p. 837), P = 5C,whi?. Assuming h, = 0. 9h, 
the formula changes to 


P=3 X C,w(0.9h)? = 0.405C,wh?. 


The bending moment per foot of length of wall, 
hp ke tex + 0. 1h)P =e 0.42050 wh? =" OF162C wh. 


The pressure on foundation due to the bending moment, 


D1 = bpp = £0,972 wh Wa 
Plus should be used for the toe and minus for the heel. 

The vertical pressure on the foundation consists of the weight of 
earth and the weight of the wall. The weight of the toe may be neg- 
lected. ‘The weight of the portion abcd may be considered as equal to 
the area multiplied by unit weight of earth. The weight of concrete 
is larger than that of earth, but the assumption is accurate enough for 
practical purposes. To compensate for greater weight of the base, 
the height may be increased by, say, 10 per cent. Thus the weight of 
the rectangle is (using same designations as in Fig. 299, p. 844) - 


W = wil — m)b X 1.1h. 
The distance of the center of gravity of this mass from the edge equals 


aa —m). Thus the pressure is applied eccentrically on the founda- 


* The principle is same as applied to eccentrically applied loads on p. 169. 


848 RETAINING WALLS 
b b 

sl — m) = Gli cae 
With this eccentricity, the pressure will be distributed on the base 
as follows: 


tion and the eccentricity 1s ; — 





m 
i eae 
ot ae + es an + Bm| =" we m)1.1A{1 + 3m]. 
Pressure at toe is 1.1wh(1 — m)(1 — 3m). 
Pressure at heel, 1.lwh(1 — m)(. + 3m). 


Combining the pressures due to earth pressure with the Snover pres- 
sures, we have 
Total pressure at toe, 


3 
py = Llwh(1 — m)(1 — 3m) + 0.9720 0s 


I 


2 
wh {1.10 TC aes 0.9720, (5) | 
Total pressure at heel, 


p2 = 1.lwhQl — m)(1 + 3m) — 0. 972C 0h 


i wh [1.10 a) (ae Bre 0.9720,(7) |. 


Resultant at Middle Third.—Assume that it is desirable that the 
resultant strike the base at the middle third. For such a condition, 
the pressure at the heel is zero. Then, 


p. =0= wh |. 1(1 — m)(1 + 3m) — 0.972C, (5) | 


Solving this for z we have, 











b | 0.972C, ‘oe 
hie SAV 1(1 — m)(1 + 3m) = 0.94) —m)(1 + 3m)’ 


, Co E 
b= 0.944) Semen 


If it is desired that the resultant strike the base for earth pressure 
multiplied by ratio 7, value of rC, should be substituted in the formula 
fore’ »: 


or 








CANTILEVER WALLS 849 


Values of C, may be obtained for different values of Cheor ron 
Resultant in the Center of Base.—If it is desired to get the resultant 
in the middle of base, then pi = po, 


wh] 1.1 — m)(L = 8m) + 0.972¢,(). | a 


= wh [1.101 = m)(1 + 8m) — 0.972¢,(5) |, 


2 
2X 0.972C,(5) = 1.1(1 — m)[1 + 3m — 1+ 3m], 


h\? _ 6.6m(1 — m) 
(; ~ 2X 0.9726,’ 


b 20: wees a 
om = ee Can PRON nce 


For ordinary conditions, when m = i Ge nO pe 








CANTILEVER WALLS 


Description.—A cantilever wall, in general, consists of an upright 
cantilever slab, retaining the wall, and a cantilever base slab, dis- 
tributing the pressure on the foundation 


AN YE ES C/o 
; Ne ar ees ) = 

ey oy ae 

ba 1-0 . 

0") Kee 





Fig. 300.—Types of Cantilever Walls. (See p. 849.) 
Nore: Reinforcement of wall not shown. 


. The 'T-shaped Wall resembles an inverted letter T. The upright 
slab is placed some distance from the toe, as shown in Fig. 300. 
This is the most commonly used type. 

The L-shaped Wall resembles a letter L. The upright slab is 
placed at the toe edge of the base; as shown in Fig. 300, with earth 


850 RETAINING WALLS 


upon the top of the base. Such a design should be used only when 
the wall is placed on the property line and it is not permissible to 
extend the toe beyond the line. 

The —\-shaped Wall is seldom used, and only for small walls in 
connection with other structures. It is uneconomical, as there is 
little earth on top of the base. 

Design of Wall.—The general proportions are assumed as 
explained on p. 845. The proportions are checked by drawing the 
resultant and finding the unit pressures on the foundation. When 
these are satisfactory, the various parts of the wall are designed for 
strength. In designing the various parts, the net expected earth 
pressure (not multiplied by factor of safety) should be used, as the 
factor of safety has been provided in selection of unit stresses 


meee eae ae aa ee ed 


Nf f EN f= 


4__Tension 
ft Crack 











¥/ 
Crack 


Fig. 301.—Failure of Component Parts of Cantilever Wall. (See p. 852.) 


= == = 7 WS 


ompression 





‘Excessive C -- Tension 


Upright Slab.—The upright slab is a cantilever loaded by the 
earth pressure, and supported at the junction of the slab with the 
base. It may fail by cracking at the inside face, by crushing of the 
concrete at the outside face, or by pulling out of the reinforcement. 
(See Fig. 301 above.) 

The upright slab does not need to be designed for the same earth 
pressure for which the wall, as a whole, is designed. In Fig. 298 the 
triangles of earth pressure are shown for level ground and for sloped 
ground. The pressure acting on the wall as a whole is indicated by 
P with a maximum unit pressure p. The pressure to which the 
upright slab is subjected is indicated by P1 with a maximum unit 
pressure pi. For level eround, Fig. (a), the earth pressure for the up- 
right slab is somewhat smaller than for the wall as a whole, the differ- 
ence being due to the difference in height. For sloped ground, Fig. (6), 


CANTILEVER WALLS 851 


there is considerable difference between the earth pressure acting on 
the wall, as a whole, and that for which the upright slab is designed. 

The maximum bending moment at the bottom of the upright 
wall is determined by the formulas that follow: 


Let h2 = distance from top of wall to junction slab and base, ft.; 
h3 = height of surcharge, ft.; 
C, = constant for earth pressure, table, p. 837; 
w = weight of earth, lb. per cu. ft.; 
6 = angle of slope of ground. 


Then 
Maximum Moment, Level Ground, 
Pepe ne abbot 2Cpwhs*in.-Ib, oP PPE 16) 
Maximum Moment, Wall with Surcharge, hs, 
M = §C,whe2?(he + 3h3) ft-lb. or 2Cy,wh2?(ho + 3h3) in.-lb. (17) 
Maximum Moment, Sloped Ground, 
M = §C, cosé whg’ ft.-lb. or 2C,cosd whe? in-lb.. . . . (18) 


_ Moments at intermediate points may be obtained by substituting 
in the above equation, for hz, the depth of the point under considera- 
tion below the top. 

After the bending moments have been computed, the thickness 
of slab and the areas of steel are computed from ordinary slab formu- 
las. The effect of the weight of wall and of the vertical component 
of earth pressure (if any) may be neglected. They increase the 
compression stresses and decrease the tensile stresses. 

The bending moments decrease toward the top. This permits a 
reduction of thickness of slab and in amount of reinforcement. 
Usually, a thickness at the top is selected by judgment (8 to 12 in.) 
and a uniform slope is adopted for the wall, as governed by this 
minimum thickness on the top and the computed thickness (or 
“depth ”’ considered as a slab) at the bottom. 

In addition to the reduction of the thickness of slab, as explained 
above, the amount of steel may be reduced. This is usually accom- 
plished by not extending all of the bars to the top. Some of them 
may be stopped forty diameters above the point where the bending 
moment is sufficiently reduced to permit the reduction in reinforce- 
ment. When it is not desired to compute moments at intermediate 
points, one-half of the bars may be stopped halfway up the wall. 


R52 RETAINING WALLS 


The bars must be anchored at the bottom by extending them 
into the base a sufficient distance to develop their strength by bond. 

Toe Cantilever (in front of Upright Slab).—The tee cantilever 
is subjected to the upward soil resistance pressure and to the down- 
ward dead load of the cantilever and the weight of earth upon it. 
Usually, the two reducing factors are neglected and the cantilever 
designed for upward soil pressure only. 

The toe cantilever may fail by cracking at the bottom, crushing 
of concrete on top, or pulling out of reinforcement. (See Fig. 301, 
p. 850.) To prevent failure, the dimensions should be computed 
according to the formulas for slabs in the chapter on Reinforced Con- 
crete. The reinforcement must be anchored by extending far enough 
beyond the junction of toe and wall. The length of imbedment must 
be sufficient to develop the full strength of the bar. 

The maximum bending moment acts at the edge of the wall. — 
The pressure is represented by a trapezoid. 


Let (in addition to notation on page 844), 
p1 = maximum pressure at toe, lb. per sq. ft.; 
p2 = minimum pressure at heel, lb. per sq. ft.; 
b = width of base, ft.; 
m = ratio of length of toe to width of base, b; 
mb = length of toe, ft. 


Then 


Maximum Bending Moment, Toe Canitzlever, 


I 


I 


M = 4[p1(8 — m) + mpez|(mb)? ft.-lb. 
or ; (19) 
2[p1(3 — m) + mpe2|(mb)? in.-lb. 
When the pressure at the heel is zero, then pe = 0 and 
M = ipi(3 — m)(mb)? ft.-lb. 
or |. (20) 
2n1(3 — m)(mb)? in.-lb. . 


Maximum Shear at Edge of Wall, 
Pil aM) salt 
V= 9 mb: thnk eee 


When pressure at the heel is zero 


Va ie amb. ise 


CANTILEVER ‘WALLS 853 


Heel Cantilever (Back of Wall).—The heel cantilever is subjected 
to the weight of the earth above it and, in case of inclined earth pres- 
sure, to its vertical component. These downward pressures are 
partly offset by the upward reaction of the soil. The bending moment 
due to downward loads should be computed separately, and from it 
should be subtracted the separately computed moment of the 
upward pressures. 

The heel cantilever may fail by cracking at the top, at the junction 
between the wail and cantilever, by crushing at the bottom, or by 
pulling out of the reinforcement. (See Fig. 301, p. 850.) To guard 
against the first two classes of failures, the depth and the amount of 
steel must be computed according to formulas for reinforced concrete. 
To guard against bond failure, the ends of bars must be anchored in 
the wall, and the bond stresses in steel must be within working limits. 


Let n = ratio of length of heel to width of base; 
nb = length of heel to inside face of wall, ft.; 
h = height at inside face of wall, top to base, in ft.; 
h, = height at edge of toe, top to one foot above base, ft. : 
hg = height of upright wall above top of base slab, ft.; 
p1 = maximum pressure at toe, lb. per sq. ft.; 
pz = minimum pressure at heel, Ib. per sq. ft. 


Shears on Heel Cantilever for Level Ground.— 
Downward Shear,‘ 


Me MohniDi cr a Bee ry (23) 


Upward Shear, 





ee Es ip, Bilas fu (24) 


Resultant Shear to be Used in Design Acting Downward, 
Reale a fay tD. dP minh er ed OR (D5) 


Moments for Level Ground.—When the ground is level, as in 
Fig. 299, the bending moments acting on the heel cantilever are: 
Downward Moment Due to Weight of Earth and Slab,!° 


My, = 0.55wh(nb)? ft.-lb. © 
or . (26) 


6.6wh(nb)? in.-lb. 


1 Depth, h, was multiplied by 1.1 to compensate for the use of a larger unit 
weight of concrete than the unit weight, w, used in the formula. 


R54 RETAINING WALLS 


Upward Moment Due to Soil Pressure, 


Me = 3[p2(3 — n) + npi|(nb)? ft.-lb. 
or hdtaediannen it xy: 
Q[n2(3'— n) + np.|(nb)? in.-lb. 


For total moment, deduct M2 from M,. Thus 
Resultant Moment to be Used in Design (Acting Downward), 
M = Mim Move ce) cen ae eee (28) 


Shears on Heel Cantilever for Sloped Ground.— 
Downward Shear Due to Weight of Earth and Slab, '° 





V1 = 0.55(h + hy + 1)wnb Ibs ie (29) 
Downward Shear Due to Vertical Component of Earth Pressure, 
V3 = ie 3 hac sinétanédwnb lb. . . - - (30) 


Upward Pressure of Foundation, 
Vo =!a{pe(2 — 0) a npy abd: beso = ae (31) 
Resultant Shear to be Used in Design (Acting Downward), 
V=Vi4t+ V3 — V2. 


For small angles, 5, value of V2 is negligible. 
Bending Moments on Heel Cantilever, Sloped Ground.— 
Bending Moment Due to Weight of Earth and Slab, 


My = Ltw(2hs + h + 2)(nb)? fold. 


or 3 Cz 
2. 2w(2hi + h + 2)(nb)? in.-lb. 
Bending Moment Due to Vertical Component of Earth Pressure, 
Mz = 4w(2h, + he)C, sind tan 6(nb)? ft.-Ib. . | 
or | (33) 
2w(2hi + h2)C> sind tan (nb)? in.-lb. 


11 The vertical component of the pressure, represented by abcd in the triangle 
of pressure, acts on the heel cantilever, as shown in Fig. 302. The vertical 
component of the earth pressure at the heel is Cpwh sin 6, and at the edge of 
upright slab Cpwh, sin 6. To get vertical components for unit of length of base, — 
these values must be multiplied by tan 6. Thus, at the heel, the vertical com- 
ponent per lineal foot is Cpwh, sin 6 tan 6, and at the wall Cpwhz sin 6 tan 6 


CANTILEVER WALLS 855 


The formula for upward moment is the sameas for level ground. 
= tpl — n) + np] ft.-lb. | 
ere ge OA 
2[p2(38 — n) + npi|(nb)? in.-lb. | 
Resultant Moment to be Used in Design (Acting Downward), 
en) 2 Nk vss ce vw we se, (OD) 


For small angles, 5, the value of M3 is negligible. 


Moe 


or 





a 
4 
ana Seng 2 
1 i bic 
' 1 ae ' ee 5h 
: 3: 1 C, d, cd Cpwh x 
eae ; 
4 y 1 
' 1 aS 
' uy eS 
r) ' is 
ecad 
) 4 L} 
: ! bio 
a & | (abe,d,= 
t * || Earth Pressure 
cael Fs ~==~12 on Heel Cantilever 
h. RY pe en = 5 
ae: b || c,d ‘ab’= Vertical 
; tS : qs Component 
4 , cae ‘ c CP . i} ) 
' pON a 
"4 aa BHC a 
ea nt 
Stele es ale b ie Se 


== 
ee Be 


s gg Ps: Upward p, 
Upward Pressures, -- a on Heel Conte 
on Toe Cantilever eS 
~np, +(1-n) Py 


1-m) Pt mp, 
Fia. 302.—Forces Acting on Base Cantilevers of T-wall. (See p. 854.) 


) 


Temperature Reinforcement.—To prevent open vertical cracks 
in the wall, which are unsightly, the wall, even if well reinforced 
horizontally, should be built in sections not more than 60 ft. long. 
Grooved lock-joints should be provided at the ends of the sections. 
The amount of longitudinal temperature reinforcement necessary 
will depend upon the degree to which elimination of cracks is desired. 
It should not be less than 0.2 per cent of the volume of the concrete. 
For water-tight walls, this should usually be increased to 0.4 per 
cent. Two-thirds of the temperature reinforcement may be placed 
near the outside face, and one-third near the inside face of the wall. 


856 RETAINING WALLS 


Adjoining bars should be properly lapped, to develop their full 
strength by bond. Deformed bars are more effective than plain 
bars in distributing the fine cracks that may occur. 


EXAMPLE OF T-SHAPED RETAINING WALL 


Example 1.—Design a retaining wall, 12 ft. above ground, to support a sand 
fill, which slopes upward at a rate of 1 to 2, making an angle with horizontal of 
26° 30’.. Weight of sand is 100 lb. per cu. ft., and its angle of internal friction 
35°. Usel:2:4 concrete and mild steel with ration = 15 and working stresses 
in Ib. per sq. in., fc = 650, fs = 16.000, v = 40, wu = 80, for plain bars, and 
u = 100 for deformed bars. 

It is desirable that the resultant should strike the base at the edge of the 
middle third for not less than 14 times the expected earth pressure. Thus, ratio 
Teli Be 

Solution.—If the base is imbedded 4 ft. below the surface, to protect it from 
frost, the total height of wall from base to top is 16 ft. 

Proportions of Wall.—The depth of wall is 16 ft. A ratio of length of toe to 
base, m = 0.3, will be assumed. For an angle of repose ¢ = 35° and an angle 
of slope 6 = 26° 30’, the earth pressure constant, from the table on p. R37 18 
C, = 0.38. The value of rCy cos 6 = 1.5 X 0.38 X 0.894 = 0.51. For this 
value and from m = 0.3, the constant from table on p. 845 is C, = 0.58, and the 
length of base 


b = Cyh = 0.58 X 16 = 9.3 ft. Use 9 ft. 4 in. 


From this the length of toe, mb = 0.38 X 9.3 = 2.8 ft) “Use 2 ft. 10 in. 

Check of Assumed Base Width.—The assumed proportions will now be laid 
out to scale as shown in Fig. 303. To facilitate computations, the concrete 
slabs are represented by rectangles. The thickness of the upright wall and of the 
base is assumed as 12 in. The weight is now computed. Taking unit weight of 
concrete at 150 lb. and of sand as 100 Ib. per cu. ft., 


Distance from Heel to -Moment of 


Weight per Foot of Length ' Weight 
Center of Gravity oak Elen! 
Slab W,=15.0 1.00 X150 = 2 250 lb. 5 ft. 6 in. +6 in. =6.00 ft. 13 500 ft.-lb. 
9.33 
Base We= 9.33 1.00 X150 = 1400 |b. aie Pee ft. 6 550 ft.-lb. 
15.0+17.7 5.517.7+2X15.0 
Ve= §5.5———— X100= 9 000 lb. ee Se tt ek Oe 
BET Ae Sorry 3° 17 7-Eib0 


Total weight.occ ones sys os os W = 12 650 lb. Total moment... ...----- 44 150 ft.-lb. 





Distance from heel to center of gravity of all loads, equal to total moment 
divided by total weight, is 44156 = 3.5 ft. 

The earth pressure acting parallel to slope is represented by a triangle. It is. 
assumed that the pressure on the face of the base is resisted by the passive pres- 





857 


EXAMPLE OF T-SHAPED RETAINING WALLS 


: NY 6 
4 006 GLISJ m0 J02H2A —-g, nog z= 4 M4 M 


\ 


TReee Vomee M 


107s 
i 


Fe Be ROLES 


! 
' 
S 2 Se 
oe $3 \a\s 
; : 3a \O 
e, —& iret \ 

94 068 NY ay rurynsoy fo 91 000 6="M—~S_ : 











---- 5 Ft, 44------ 





a 
91 €6 eS 
: ' i 35 
4 xy 
t i] t ~ 
: ek | oR 
| S| aS 
' I Se) 22 
al | : Spa) 
Net 23 
Ea | = 
Ww ' L ~ 
© | 1 
‘ n 
‘ 1 
' ‘ 
1 ' 
1 i] 
' I 
\ ! 


‘QI LGES 





Se 
> 
2% 
uN 
iS 

® 
ag 
D Fae 
Ss 
Sw 
Q 
2150 
=m™ 
™ 


1 OLS & 


Pressure on Foundation for Expected Earth Pressure 


| Pressure on Foundation for Earth Pressure X Yr 


(See p. 856.) 


Fig. 303.—General Dimension of Wall. 


858 RETAINING WALLS 


sure at the toe. The total earth pressure, from Formula (4), p. 837, where 
C, 018318 =x 0.38 X 100 X 17.72 = 6000 Ib. This applies at one- 
third of the height from bottom. 

The components of the earth pressure are: 


H = P cos 6 = 6000 X 0.894 = 5 360 lb. 
V = Psin 6 = 6000 X 0.446 = 2 680 lb. 


Plot the resultant downward force, W, at a distance of 3.58 ft. from the heel, 
and also the earth pressure, P, in proper position. Combine the two forces as 
shown in the figure. The resultant, R, strikes the base well within the middle 
third. 

Combine again the force, W, with rP, i.e., for earth pressure multiplied by 
ratio r = 1.5. For this condition, the resultant strikes the base at the edge of 
the middle third. 

Pressure on Foundation.—The pressure on the foundation is obtained from 
Formula (14), p. 847. The vertical component is 15 330 lb. and the eccentricity, 








; 9 ft. 4 in. 5 
e = 5 ft. 41in. Eas = Sin. = 0.67 ft. 
15 250 6 X 15 330 X 0.67 
= = 1:645 +) 712 1b, a tte 
Hasne PP oe 9.33? + 712 Ib. per sq. 
From this, 


pi = 2357 lb. per sq. ft.; p2 = 933 Ib. per sq. ft. 


For the other condition, the pressure acts praetically at the edge of the middle 
third. Therefore, the maximum pressure at the toe equals twice the average 


pressure. 


—7 2 pS es 3 5 TO) b. e oq. nN. 


Graphical method of determining pressure on foundation for known average 
pressure and location of force, is shown in Fig. 303, p. 856. 


Resistance to Sliding —The horizontal component of the earth pressure is 


5 380 
= 0.352 
15 330 — 


is much smaller than the tangent of the angie of friction, hence the design is 
satisfactory. (See p. 839.) 

Toe Cantilever.—For explanation, see p. 852. 

The maximum bending moment, from Formula (19), p. 852, when m = 0.3; 
mb = 2.82 ft py ="2 257 1b a 933 Ib. 


M = 2|2 357(3 — 0.8) + 0.3 X 933]2.827 = 106 000 in.-Ib. 





H 
H = 5380 lb. The normal resultant is N = 15 330. Ratio N ae 


The maximum external shear, from Formula (21), p. 852, 


2 357(2 — 0.38) +0.3 X 933 
= ) f 0.8 Ae" 2.82 = 6050 Ib.) 


y. 
2 





EXAMPLE OF T-SHAPED RETAINING WALL 859 


The depth of slab for bending moment, from formula d = C,;V M, where 
C;, = 0.028 (Table 2, p. 880). 


d = 0.028V106 000 = 9.12 in. 
As = 0.077bd = 0.0077 X 12 X 9.12 = 0.84 8q. in. 


3 bars, 4’ on centers has an area of 0.921 sq. in. 


Bond Stresses now will be investigated. The perimeter of a 2-in. rd. bar is 
o = 1.96. Since there are 3 bars per lineal foot, the value of So = 3 X 1.96 = 
5.88. 

The unit bond stress is 


V 6 050 


ee eee 196 Ib: 
Lojd 5.88 X 0.875 X 9.3 


The bond stress is excessive. To reduce it, increase either the area of steel or the 
depth. In this case, the depth will be increased in the ratio of the computed 
to the allowable bond stress. 


d, = 9.12 X 128 = 11.5 in. 


Use a total depth of 15 in., which gives about 3 in. protection for the steel. 
The bars must be extended 40 diameters beyond the junction of the toe and 
the wall, to develop their full strength. For 3-in. bars, this amounts to 2 ft. 1 in. 
Shear, 
6 050 . 
= = 48 0 Ib. +1h; 
* ~ 11.6 X 0.875 X 12.2 AG tan Tt 
Upright Slab.—For explanation, see p. 850. As evident from Fig. 305, p. 862, 
the height of upright slab above the junction of slab and toe is hy = 14 ft. 9 in. 
Since Cp = 0.38 and w = 100 Ib. per sq. ft., 6 = 26° 30’, cos 6 = 0.894. The 
bending moment, from Formula (18), p. 851, 


M = 2 X 0.38 X 0.894 * 100 X 14.753 = 218 000 in.-lb. 
Thickness of slab, from formula d = C iv M is 


d = 0.028V 218 000 = 13.1 


Use h = 15 in. With 13 in. protection outside of bars, the distance to the 


0.7 : 
center of steel is 1.5 + “e ='] 875 in. 


Area of steel, from formula A, = pbd, where p = 0.0077, is A; = 0.0077 X 
12 X 138.1 = 1.21 sq. in. This is satisfied by 3-in. rd. bars placed 4.37 in. on 
centers. Use 44-in. spacing at the bottom. 

At the top, where the bending moment is zero, the minimum practicable 
depth may be used. This is considered to be from 8 to 12 in. In this case, 
10 in. depth is selected at the top and the slab made to slope uniformly to the 
- Maximum depth. 


860 RETAINING WALLS 


If it is desired to make some of the vertical bars shorter than the rest, 
bending moments at intermediate points should be found, and the required 
areas of reinforcement computed. Since the moments vary with third power of 
the height, the values at intermediate points may be obtained by multiplying 
the maximum moment by third power of the ratio of the height of the point to 
the total height of upright slab. 














Effective M 

Height hz | Ratio, iis 's) See al Depth, Aye fad 
h k nch-pounds eo fsJ 
nches eq. In. 
Pit Os 1 o 3 400 9.35 0.03 
ih = 188 4 $ 27 200 10.6 0.18 
8h = 11.05 A 27 §2 000 11.85 0.56 
he = 14.75 1 1 218 000 13.1 121 








; ee ee 
For maximum bending moment use 3.in. rd. bars spaced 4% in. on centers. 


The length of bars may be obtained by drawing a free-hand sketch on cross- 
section paper, aS shown in Fig. 304. The height is laid out to a convenient scale, 
for instance, one division for one foot. At each elevation, the required area of 
steel is laid out on a horizontal line, the divisions being used again as a measure. 
The points connected give a curve representing the required area of steel at any 
height. The reinforcement is laid out as used. In the sketch, an assumption 
was made that one bar in six would be carried to the top, making the top spacing 
of bars 25} in. Next, the spacing of bars is reduced to 123 in. by using two bars 
in each 253 in. Then four bars are used in 254 in., and finally six bars, making 
the bottom spacing 44 in. The areas of bars are reduced to effective areas per 
lineal foot of wall, and plotted. ‘Thus, one 3.in. rd. bar in 253 in. gives 0.207 
sq. in. per lineal foot, two bars 0.41 sq. in. per lineal foot. The intersection of 
the vertical lines representing the areas of bars, with the curve representing the 
required areas of reinforcement, gives the theoretical point where the bars become 
unnecessary. The bars should be extended 40 diameters beyond this theoretical — 
point. In important work, hooks should be provided at ends to prevent slip of 
bar. 

Heel Cantilevers.—For explanation, see Pp. 953. Since nb = 5 ft. 6 in, 
n= a = 0.59, p: = 2 357 |b., po = 933 lb., A = 16 ft. Oin., ki = 17 ft. 9 ing 
h, = 15 ft., sin 6 = 0.446, tan 5 = 0.499, the maximum bending moments are — 


as follows: 
Bending moment due to weight of earth and slab (Formula (32), p. 854). 


M, = 2.2 X 100(2 X 17.75 + 16.0 + 2)5.5% = 357 500 inl. 


Bending moment due to vertical component of earth pressure, Formula (34), 
p. 855. 


M, = 2 X 100(2 X 17.75 + 15.0)0.38 X 0.446 xX 0.499 x 5.52 = 25 900 in.-lb. - 


EXAMPLE OF T-SHAPED RETAINING WALL 861 


The maximum bending moment due to upward pressure, from Formula (33), 
p. 854, is 


M; = 2(933(3 — 0.59) + 0.59 X 2 357]5.52 = 221 000 in.-lb. 
The sum of the bending moment is 


M = My, + Mz — M; = 357 500 + 25 900 — 221 000 = 162 400 in.-lb. 

































































































































































































EE 1 Ee 

: aC In06 1000 

wah mE 
: Ce : 
RE 
| i-He -5 il 
ae ee raphe 
aan d dnd used | to oh A 
BEE vealFoot | | | 
ole oe Wl 
HHH ‘ai use| V-/4t6{ |_| 

[) 
meen Pees 
SCH E Ee Bert heh at wl ETA 
= el 








Fia. 304.—Points of Stopping at of Vertical Bars in Upright Slab. (See p. 860.) 


Thickness of slab, computed as before, 


d = 0.028V 162 400 = 11.3, 
ee) 3 te k= 13-28. Usa 132 in 
A, = 0.0077 X 12 X 11.3 = 1.04 sq, in. 


Use 3-in. rd. bars, 5 in. on centers, with an area of 1.06 sq. in. 

Temperature Reinforcement.—In the upright slab, temperature and shrinkage 
reinforcement will consist of horizontal bars placed near both faces of the slab. 
The total amount will be assumed as equal to 0.2 per cent of the total concrete 
area. This requires 12 X 10 X 0.002 = 0.24 sq. in. of steel per foot of height 
at the top of the wall, and 12 X 15 X 0.002 = 0.36 sq. in. per foot of height at 


862 RETAINING WALLS 


the bottom. Of this steel, about two-thirds will be placed near the outside face 
and one-third near the inside face, as shown in Fig. 305. To support the bars 
near the outside face, either spacers or vertical bars should be used. 


Design of L-walls.—L-walls consist of two instead of three 
cantilevers, as the toe cantilever is omitted. The method of design 





S 
—— 


+= 





















| 1 
eg 
! 1 \ 
| | \ { = 
ns ie Eh 
eed is <|~~s. All longitudinal bars & 
© " . 
| | <4 > a 3 rd, continuous = 
ea ~ with 40 diam. splice. Z 
' ) N = GC 
ie be ? 
{ | Is | aan 
ae i \ Be 
! 1 I Arad 
a | i a | MM 
Seve igi 2 
3 Me: 
cc 8 : FS 
Pa wig | oD 
t 
ee } : + 
j ~ ; oO w 
i bas vy 
Ea = | 
ro \ 
vo 
ee 
i | 
! \ 
: | 
Yi) Y 
Bis 
| 
i 3] 
i] Ss 
1 2. 
[aig 
S 
t I | = 
! Py “8 
t bey 2: * I 5 
iad 4"on 2 Ft. 7 { ws 
8 centers | a"chorage Construction 
1-1 Ft. 3” Joint 
-2 Ft, 10'->}< E><——- - 5 ft, 3 —--- 
------ == — 9 Ft, ¢’-------- 


Fic. 305.—Details of T-wall designed in Example. (See p. 862.) 


of the upright slab and the heel cantilevers is the same, as explained 
in connection with T-beams. ‘The arrangement of reinforcement 1s 
shown in Fig. 306, p. 863. 

Erection of Cantilever Walls.—In designing retaining walls, the — 
method of erection should be considered. Ordinarily, it is desirable 
to erect the base slab first, before the forms for the upright slab are 


WALL WITH COUNTERFORTS 863 


erected. The forms for the upright slab are then supported on the 

base. To avoid the necessity of bracing forms on the bottom, a 

small section of the 

wall, say, 6 in. high, 

is built with the me 

base. The wall eo UE 

forms then are bolt- ate 

ed to this stump. 3 
The wall rein- Al 

forcement extends OOM TG sA 

d ; Reinforcement 

into the foundation 


slab. There would as 

be considerable diffi- eI) _--- Main Tension Reinforcement 
culty in keeping the 3: HOB OIRO Lo 
long vertical bars es 

in position during Re: 

pouring of the foun- ae 

dation slab. To  . soe 


avoid this, dowels = aN a 
are inserted instead OF riction bq] Dowels 
of full-length bars. Joint --- 7 PE 
One-half of each ; 
dowel is imbedded ae 
in the base and the nome le 
other half extends Fic. 306.—L-shaped Retaining Walls. (See p. 862.) 
above the construc- AYR 
tion joint to lap the tension reinforcement of the upright slab. 
The length of dowels is such that each dowel is capable of developing, 
by bond, full strength of bar below and above the joint. 

To transfer shear at the construction joint, a key should be used 
as shown in Figs. 305 and 306. 


,Main Tension Reinforcement 
for Base Cantilever 






BS a= 28. 
Deis Ceo Slot ia ans 


WALL WITH COUNTERFORTS 


To reduce the amount of steel and concrete in the upright slab 
and the toe cantilever, a design is often used with a counterfort 
between the two cantilevers, as shown in Fig. 307, p. 864. 

Relative Economy of Cantilever Walls and Walls with Counter- 
forts.— Whether the T-section of reinforced wall or the wall with 
counterforts is the more economical depends upon certain conditions. 


864 RETAINING WALLS 


The principal condition is the height of the wall, but the intensity 
of the earth pressure and the relative cost of concrete and steel and 
forms also enter into the 
consideration. The T- 
section is preferable where 
skilled labor is scarce, as 
the construction is sim- 
pler and the placing of 
steel easier. The form 
st construction in the coun- 
8 terforted wall is consid- 
< erably more expensive. 
Comparative studies of 
the two types indicate 
that the counterfort type 
is scarcely ever economi- 
cal when the height is 
less than 18 feet. 

For lower walls, it 
will be found that the 
\ additional cost of form- 
Fig. 307.—Isometric View of Wall with Counter- work for the counterfort 

forts. (See p. 863.) more than balances the 
saving in material. 

Design of Wall with Counterfort.—General proportions of the base — 
will be the same as for a cantilever wall of the same height and sub- | 
jected to same earth pressure. The proportions should be checked, in 
the same manner as shown in the example for cantilever walls, p. 857. 

Stresses Acting on a Wall with Counterforts.—An idea of the 
distribution of stresses in a wall with counterforts may be had from 
Fig. 308, which shows the possible causes of failure of such wall. 

Figure (a) shows a failure caused by cracking of the counterfort. 
This occurs when the amount of tension reinforcement near the out- 
side face of the counterfort is too small. 

Figure (b) shows a failure caused by the separation of the upright 
wall from the counterfort, which is caused by insufficient anchorage 
of the wallto the counterfort. Since the upright slab is reinforced with 
horizontal bars, it cannot act as a vertical cantilever, as in T-wall. 
Separation of the upright wall from counterfort, therefore, means 
failure of wall. 





Upright Slab-.. 


--Counterfort 


ace 


WALL WITH COUNTERFORTS 865 


Figure (c) shows failure of the upright slab produced by earth 
pressure. ‘This is caused by insufficient amount of tension reinforce- 
ment. 

An additional type of failure may occur by rupture of the toe can- 
tilever. This is not illustrated. It is similar to that shown in Fig. 
301, p. 850, in connection with failures of T-wall. 

At the junction of the wall and the base, the slab is not free to 
defect horizontally, as it is held by the base. A vertical bending 
moment will develop there, which should be resisted by short vertical 
bars placed near the outside face of the wall. 















Direction of 


\ Direction of | Earth Pressure 


-\ Earth Pressure 
\ ce 


we — i, 
Sy Ete. 
ee as 


Horizontal Section through Wall 


Failure of Upright Slab 
\ Crack in ae 
—~Counterfort 
ge 4 ane si ae D.. 
Failure of Counterfort Upright Slab Separated 
from Counterfort 
(a) (b) (c) 


Fig. 308.—Failure of Component Parts of Wall with Counterforts. (See p. 864.) 


Upright Slab in Wall with Counterfort.—The upright slab is a con- 
tinuous slab loaded by earth pressure and supported by the counter- 
forts. The principal reinforcement, therefore, is horizontal. In 
designing, consider horizontal strips of wall, say, one foot wide. 
The loading on each strip will be uniform and equal to the intensity 
of earth pressure acting at the height of the strip. The bending mo- 

2 
ment should be figured by formula M = oe for interior panels, and 
2 
ay we 


10 for end panels. 


866 RETAINING WALLS 


The earth pressures at the various strips varies from a maximum 
at the bottom of the slab to zero at the top. The thickness of slab 
is computed for the maximum pressure at the bottom of the slab. 
The slab may be made of uniform thickness. However, if the differ- 
ence between the maximum thickness and the practical minimum is 
large, the wall may be sloped. 

When the thickness of the slab has been decided upon, the amount 
of reinforcement required at different heights should be computed. 
Since the unit pressure varies according to a straight line from zero 
at the top to a maximum at the bottom, the bending moments also 
will vary in the same manner. If the heights are plotted on a 
vertical line and the moments at the bottom on a horizontal line, the 
triangle will represent moments at various points. These may be 
scaled. 

Since the wall is a continuous slab, reinforcement should be pro- 
vided at the supports. The reinforcement in the middle of the span 
will be placed near the outside face of the wall, and the reinforcement 
at the counterforts near the inside face of the wall. 

Toe Cantilever of Wall with Counterfort—The method of 
design of the toe cantilever is the same as for cantilever walls. The 
formula given on p. 852 may be used for bending moments and 
shear. 

If the length of the toe cantilever is appreciable, some saving may 
be effected by using cantilever ribs spaced at the same intervals as the 
counterfort. The base slab may then be considered as spanning 
between the ribs. 

Horizontal Slab Back of Wall.—The horizontal slab forming the 
base of the wall may be considered as though it were composed of 
continuous strips parallel to the wall, supported by the counterforts. 
Each strip is uniformly loaded, but the loading of the various strips 
of the slab varies from a maximum at the edge of the base to a 
minimum at the wall. For level ground, the loading of each strip 
is composed of the weight of the strip and of the earth above it, 
minus the upward pressure at the foundation. 

For inclined ground, the downward load consists, in addition to 
the weight of earth and slab, of the vertical component of the earth 
pressure. If 6 is the angle of slope, C, the earth pressure constant, 
hy the distance from top of slab to top of ground at the end of base 
in feet, and hz the height of wall above base in feet, then the intensity 
of the vertical component of earth pressure equals: 


WALL WITH COUNTERFORTS 867 


Vertical Component of Earth Pressure at End of Base, 


Nmax = C,pwh, sin 6 tan 6 Ib. per sq. ft.. . . (86) 
Vertical Component of Earth Pressure at Wall, 
N = C,whg sin 6 tan 6 Ib. persq. ft.. . . (87) 


The loading is represented by a trapezoid, from which the load 
at any point may be computed. 

Counterforts.—Counterforts are really upright cantilever T-beams 
held in position by the base slab and loaded by the earth pressure 
transferred to the counterfort by the vertical slab. A portion of the 
vertical slab may be considered as a flange of the T-beam. The 
amount of earth pressure on the cantilever may be determined as 
follows: | 

Let ho = height of cantilever above base slab, ft.; 

s = spacing of cantilevers, ft.; 
C, = constant from table on p. 837, for earth pressure; 
6 = angle of inclination of earth pressure; 
w = height of earth per cu. ft. 
Then 
Total Pressure upon Counterfort Acting Horizontally, 


eee oS ID Pe ee eS (88) 
Maximum Bending Moments in Counterfort, 

M = 1C, cos 6 whg?s ft.-lb., or 2C> cos 6 whe3s in.-lb. (39) 
For level ground 6 = 0, consequently cos 6 = 1. 


The maximum bending moment and the total pressure act at the 
junction of the counterfort with the base slab. The pressure and 
bending moment at any other point may be found by substituting 
the proper value for he. 

The amount of steel to resist bending moment may be found in 
the same manner as for an ordinary T-beam. The bars must be 
securely anchored at the bottom by bending them back into the 
base slab. The bending moment decreases rapidly. Since the 
cantilever slopes, the available depth of beam also decreases, but 
in a lesser degree. It will be found that it is possible to reduce the 
amount of tensile reinforcement by making some of the bars shorter 
than the rest. To find where the bars may be stopped, plot the 
required amount of steel and the available amount of steel. Carry 
the bar at least 40 diameters beyond the point where, theoretically, 


868 RETAINING WALLS 


its area can be dispensed with. Because of the importance of the 
counterfort, it is advisable to provide hooks at the ends of all bars 
except the longest. 

Horizontal stirrups should be provided, not only to resist diagonal 
tension in the counterfort, but also to tie the wall to the counterfort. 
(See failure of wall, Fig.308(0).) The stirrups must be looped around 
the tensile reinforcement in the counterfort and extend through the 
counterfort as far as possible into the slab, where the ends should be 
provided with hooks. The counterfort should also be tied to the 
base by means of vertical stirrups, the area of which should be sul- 
ficient to transfer, by tension in steel, all the downward load from 
the base to the counterfort. 

Spacing of Counterforts.—The quantity of concrete and steel 
in a wall depends upon the spacing of counterforts. By reducing 
the spacing of the counterforts, the thickness of the wall and the base 
slab are reduced, hence the amount of concrete and steel is reduced. 
But, at the same time, the number of counterforts is increased, 
which means increase in formwork. The most economical spacing 
of counterforts is the one for which the total cost of materials and 
formwork is a minimum. 

Ordinarily, 12 in. is considered a practical minimum thickness 
for the vertical longitudinal slab at the bottom. Investigation shows 
that the most economical spacing of counterforts is the one for which 
the required thickness of vertical slab is equal to this minimum thick- 
ness. 

Considering 12 in. as the minimum thickness of vertical slab at 
the bottom, and using the following stresses in lb. per sq. in., fe = 650, 
fs = 16 000, the spacing of counterforts and the corresponding unit 
earth pressures at the bottom are given in the table below: 


SS ee 





8 ft. 0 in./8 ft. 6 in. 


Spacing in Feet.......-. ft. .OUne i tbegoe te 9 ft. Oin.|9 ft. 6 in.|10ft. Oin. 


Unit pressure, lb. persq. ft.) 2 000 lb.| 1 860 lb. 





1 640 Ib |1 450 Ib. |1 300 Ib. |1 160 lb. |1 050 lb. 





ee ee 


Thus, when the vertical slab is 24 ft. high above the base slab, 
weight of earth w = 100 lb., and the constant C, = 0.40, the pres- 
sure at the bottom is p = 0.40 X 100 X 24 = 960 Ib. per sq. ft: 
For this pressure, the economical spacing of counterfort is 10 ft. 

For the same conditions, except that the ground surface, instead 
of being level, is inclined 1 to 23 with the horizontal, constant 


SPECIAL DESIGNS OF RETAINING WALLS 869 


Cy = 0.60, the pressure at the bottom is 1440 Ib.; consequently, 
the economical spacing will be 8 ft. 6 in. instead of 10 ft. 

From the above, it is evident that, as would be expected, the 
economical spacing of counterforts depends not only upon its height 
but also upon the intensity of earth pressure. 

Thickness of Counterfort.—The thickness of counterfort should 
not be smaller than required by diagonal tension. The external 
shear to be used in determining the thickness is the total pressure 
from Formula (88), p. 867. 

The counterfort must be wide enough to accommodate all rein- 
forcement without crowding, preferably in not more than two 
layers. 

Erection of Counterforted Wall—The counterfort should be 
poured with the vertical wall, in one continuous operation. No 
construction joints should be made in the counterfort. In the wall, 
construction joints should be vertical and placed midway between 
counterforts. 

If the base cantilever is poured separately from the counterfort, 
concrete keys should be provided at the juncture of base and coun- 
terfort to transfer shear from the counterfort to the base. 

Design of Wall with Counterforts.—Figure 309, p. 870, shows a 
typical design of a wall with counterfort, worked out in accordance 
with the rules previously given. 


SPECIAL DESIGNS OF RETAINING WALL 


Special designs have been worked out with considerable ingenuity, 
where local conditions require departure from the standard sections. 
In railroad construction, it is often necessary to build retaining walls 
where every inch of available room up to the edge of the right-of-way 
is valuable, and where no trespassing on adjoining property is per- 
missible. 

Cellular Wall.—The simplest solution, in case of a railroad fill, 
is to use the L-shaped wall with or without counterfort. In cases 
where such a wall produces excessive pressure on the toe, the cel- 
lular type of wall, shown in Fig. 310, p. 871, may be used. It con- 
sists of a continuous base a, two parallel longitudinal walls b and 6y, 
and tie wallsc. The space between the longitudinal walls is filled 
with earth. 

The earth pressure is resisted by the weight of concrete and the 
weight of fill between the longitudinal walls. The pressure on the 


870 RETAINING 











A Iternate Arrangement 0 
Slab, Steel rf 


we 


7 
Temperature Bars 5 


All Horizontal Bars sf 



























au 
‘ 
" N 
*) 
5 
; SO. 
eb aaa 
; = Sr ee 
: mlb 
y : mie 
Ses Le 
. wey PV itros be ad [et ee a ee 
' Lol Gee 
‘ ae Ch RR ES 
: aS) WN 21 a ei 
\: re) As ee ee ee 
a Sa oe BS S| | | 
: %:=S ae 8 ia La eS BEC ii Rls Rec 
SOS aa Be! SS 
; a ara eae 
BBQLS, me 





iano 
| vase 


SS 














Cross Section Rear Elevation 
Fic. 309.—Details of Wall with Counterforts. (See p. 869.) 


SPECIAL DESIGNS OF RETAINING WALL 871 


foundation is obtained by combining the horizontal earth pressure 
with the weight of the concrete and the fill. All possible conditions 
of loading must be considered. The line of pressure may be obtained 
graphically, by the method explained on p. 856 in connection with 
cantilever walls. The distribution of pressure on the foundation 
depends upon the location of the point of intersection of the line of 
pressure with the base. It may be computed as explained on p. 849. 














is eA ati 
eave FESR SNR 
; er ys 2 Fg? 
' 4 AeA b 
' Be NAV ee Ne 
| ee ee Bea f I 
3% VA pel 3 GaN ( efarth 
Cn oe be Le SN A) ge Earth 
7 é oe oF eNO: Pressure 
Te EES CASE ee Sa XS ee 
! SF SN Wee ere rr Ne AWE ZA 
FEB Me On LSS SAI RRNESE GI] 
el \N) P \ oN : wes ‘: i 
aye CIN oes Eye: 
oat Nita St —~ 5 . 
 < Vain oa 
3 MW A; Witte ZAG S ; 
WIDE NACA 
+f Ih 
eo TS 
: WARY VARTA AN 
76' --=---- 9 Ft, 101--=----S4 ee — 16" 
= =a gt 
~—------ ih Oe 


Reinforcement not shown. 


Fic. 310.—Cellular Wall. (See p. 869.) 


For designing purposes, the longitudinal walls are considered as 
slabs supported by the tie walls and loaded by the earth pressure 
produced by the fill and the surcharge caused by the train load. 
The reinforcement consists of longitudinal bars placed near the out- 
side face of the wall, between the tie walls, and near the inside face 
at the tie walls. Vertical reinforcement should also be used, to 
prevent shrinkage and temperature cracks. The cross walls should 
be reinforced with horizontal bars of sufficient area to tie the longi- 
tudinal walls and prevent their separation under the pressure of the 


872 RETAINING WALLS 


earth in the cells. The base should be designed to resist the differ- 
ence between the weight of earth and the upward pressure of the 
soil. They may be considered as supported by the cross walls. 
Lacher Design.!2—Where the cellular type produces much pres- 
sure on the toe, the Lacher design may be used. This consists of a 





is -------- 8 Ft,- 64 ------- Z 
fg! Gree 2 j E 
ied ae = NOE 
nae =) 
fra 08 mea RITE? aN MN 
oS pe 
| | ee : D 
| Pst 
| f is 
Vetch age ls 
Dag Ee 
| fs 
(ane: 





1 Ft.- 6" Walls 


n 


| 
\ 
| 
| 4 
a Spaced 12 Ft. on centers me’ 
c cs 
2 
| Y 
| es “(| 
| s 
\ 
ie 
I . 
{ Ui 
: : 
y ou 
' sie 
! ) RU). 
: ne 
' tj 
r " a < i 
t vs <3 I 
= SS ah b 
=p W 


Reinforcement not shown. 


Fiq. 311.—Lacher Design of Retaining Wall. (See p. 872.) 


continuous base, a, longitudinal wall, 6, and cross walls, c, spaced 
12 ft. on centers, and a longitudinal beam, d. The track is sup-_ 
ported by pre-cast concrete slabs resting upon a beam, d, and a 
wall, b. The cost of the Lacher design, as estimated by Mr. Prior, 


12 J, H. Prior, in Engineering and Contracting, May 10, 1911, p. 530. 


SPECIAL DESIGN OF RETAINING WALL 873 


was larger than the cost of the cellular design. The Lacher design 
has the advantage, however, of producing lower pressure at the toe. 

Wall with Tie Rods.—An interesting wall is that designed by 
Gustave Lindenthal for the New York Connecting Railroad. Here 
two longitudinal walls, 65 ft. high, enclose a railroad fill, nearly 
60 ft. wide, carrying a surcharge loading of four tracks (EZ 60 loading) 
and 100 per cent impact. The longitudinal walls are connected by 
transverse walls about 50 ft. apart. The earth pressure produced 
by the earth between the longitudinal walls is resisted by 21-in. 
round tie-rods, spaced 10 ft. apart vertically and horizontally.18 
The ends of the tie-bars are threaded and provided with nuts. The 
pressure is transferred to the concrete by means of longitudinal and 
vertical 8-in., 163-lb. channels, which are placed near the outside face 
of the walls. The bolts at the ends of the tie-rods bear against the 
anchoring channels. While the earth filling was being placed. the 
tie-rods were surrounded with concrete to form an 8-in. square sec- 
tion, for protection against rusting. 


18'The computations are discussed in Engineering News, May 6, 1915, p. 886. 


CHAPTER XXII 
TABLES AND DIAGRAMS 


This chapter gives tables and diagrams required in design. For 
convenience of reference the tables are grouped according to the 
subject matter. 

The list of tables in the body of the book is also given. 


General 
PAGE 
Table 1.—Working Unit Stresses... 1.1... eeeee cee ecceeererccess 879 
Constants for Rectangular Beams and Slabs. 
Table 2.—Constants for Rectangular Beams and Slabs, n = 15...... 880 
Table 3.—Constants for Rectangular Beams and Slabs, n = 12...... 881 
Diagram 1.—Constants hk and p for Different Stresses f- and fs...... 882 
Diagram 2.—Value of k, j, and i for Different Ratios of Steel........ 883 
: 
Explanation of Use of Diagrams given on Pages 882 and 883. 
Design of Rectangular Beams 
Table 4.—End Spans of Continuous Rectangular Beams............ 884 
Table 5.—Interior Spans of Continuous Rectangular Beams.......... 885 


Design of Slabs 
Table 6.—Thickness of Slab and Reinforcement Required for Given 


Live Load. f, = 650, fs = 16000, n.= 15...............++-+5; 886 
Table 7.—Simply Supported Slabs, fe = *SB0ORCSSs 
Table 8.—End Spans Continuous Slabs, | fs = 16,000 22889 
Table 9.—Interior Spans Continuous Slabs, n= 15. 890 


Table 11.—End Spans Continuous Slabs, 
Table 12.—Interior Spans Continuous Slabs, 


874 


Table 10.—Simply Supported Slabs, So = (3005 2 SBE 


TABLES AND DIAGRAMS 875 


Design of T-Beams PAGE 
Diagram 3.—Values of Maximum Steel Ratios p» for 
Different Ratios 894 
Diagram 4.—Values of Constants Cq for Different { 
Ratios “ fs = 16000 894 
Diagram 5.—Values of Maximum Steel Ratios p, for 
Different Ratios 895 
Diagram 6.—Values of Constants Cg for Different 
Ratios | f, = 18000 805 
Table 13.—Values of k, j, and Cy for Different eee, 896 


Explanation of Use of Table and Diagrams given on Page 897. 
Design of Stirrups | 


ee IN OPCS ITIUIDS «eee ase §. dogoenigvie wink ected oc larrenee tS cal) 899 
Table 15.—Number of U-stirrups in Uniformly Loaded Beam........ 900 
Table 16.—Spacing of U-stirrups in Uniformly Loaded Beam........ 901 


Explanation of Use of Tables given on Page 902. 


Beams with Steel in Top and Bottom 


'( fe = 650, 750, 800, 900 
Diagram 7.—Ratios of p’ for Given 1, } f; = 16 ef | 904 
Jo =_800, a 1000 
Diagram 8.—Ratios of p’ for Given pi, } f; = 16 oe | 905 
fc = 650, a 800, 900 
Diagram 9.—Ratios of p’ for Given pi, } f, = 18 ne 906 
f. = 800, "500. 1000 
Diagram 10.—Ratios of p’ for Given 71; Jee 18 es 907 


Diagram 11.—Ratios of p’ and p; for Different at and a = 0.02 to 0.12 908 


Diagram 12.—Ratios of p’ and p; for Different a anda = 0.14 to 0.24 909 


Diagram 13.—Values of 7 for Different Steel Ratios p.............. 910 
Explanation of Use of Diagrams given on Page 903. 


- Flat Slab Design 
Diagram 14.—Constants in Formulas for Thickness of Slab at Column.. 911 


Diagram 15.—Constants in Formulas for Thickness of Slab in Middle 
EER OTIS acs Seip Ra PME RE Meld PORTE CHA toe 911 


876 TABLES AND DIAGRAMS 


PAGE 
Table 17.—Flat Slab Constants Co and C1: for Computing Compres- 
sion in Concrete: ¢. ues pas cn essen + « ein aie skenetteianenete aea 912 
Table 18.—Flat Slab Constant C12 for Computing Shearing Stresses 
at Column Head ix. ss ste se ne a 94.007) > Rape ne 913 
Table 19.—Flat Slab Constant Ci; for Computing Shearing Stresses 
at Edge of Drop Panel.........--.++ + == #0 geen i 914 
Column Design } 
Table 20.—Properties of Column Sections, Areas, Volumes, etc...... 915 
Table 21.—Average Stresses Unit f for Different Values of f-and p.. 916 
Table 22.—Average Stresses f: in Spiral Column, New York Code.... 918 
Table 23.—Average Stresses f2 in Spiral Column, Chicago Code...... 919 


Diagram 16.—Total Loads on Round Columns for Different Stresses f. 919 
Diagram 17.—Total Loads on Square Columns for Different Stresses f. 919 
Table 24.—Safe Loads, Square Columns with Vertical Reinforcement, 


f, = 450. (1:2: 4 concrete) ........ 6c eee eee eee ee eee eee. 920 
Table 25.—Safe Loads, Square Columns with Vertical Reinforcement, 
f, = 570. (1:14:38 concrete) ........ eee e eee eee eens 921 
Table 26.—Safe Loads, Square Columns with Vertical Reinforcement, . 
f, = 680. (1:1: 2 concrete)................-- +s esses eee 922 
Table 27.—Safe Loads, Round Columns with Vertical Reinforcement 
only, fo = 450. (1: 2:4 comorete)..-. 0/100 tse 923 
Table 28.—Safe Loads, Round Columns with Vertical Reinforcement 
only, fe = 570. (1: 13:38 concrete)........-..--.++2+-.20-:. 924 
Table 29.—Safe Loads, Round Columns with Vertical Reinforcement 
only, fe = 680. (1:1: 2 concrete).......... 6... sense eee 925 
Table 30.—Safe Loads, Round Columns with Vertical Reinforcement 
and 1 per cent Spiral, fe = 700. (1:2: 4 concrete)........... 926 
Table 31.—Safe Loads, Round Columns with Vertical Reinforcement 
and 1 Per Cent Spiral, f. = 890. (1: 14: 3 concrete) ....... 0 ae 


Table 32.—Safe Loads, Round Columns with Vertical Reinforcement 
and 1 Per Cent Spiral Reinforcement, f. = 1060. (1:1: 2 con- 


crete). 2600 ee a bee cece ne ei Oe 2 os eae 928 
Table 33.—Column Spiral Forming 1 Per Cent of Core............. 929 
Table 34.—Values of pA in Spiral Columns, §-in. wire............- 931 
Table 35.—Values of 1A in Spiral Columns, 7-in. wire............. 932 
Table 36.—Values of pA in Spiral Columns, 3-in. wire...........-- . 933 


Explanation of Tables 34 to 36 given on Page 930. 


Members Subjected to Direct Load and Bending 


N 
Diagram 18.—Constants C, in Formula f= Coa, for Za. hw. see 934 
: . N 
Diagram 19.—Constants C, in Formula de Coup for 2a = O:9R. =... 935 
‘ : N 
Diagram 20.—Constants C, in Formula f. = Ce—- for 2a = 0.8h 936 


Choo Ae ee 


TABLES AND DIAGRAMS 877 


PAGE 
Diagram 21.—Constants C; in Formula f, = Cm COC SLOT Ayo 937 
Diagram 22.—Values of k for Different Eccentricities, \ Both Sides 938 
Diagram 23.—Values of Ca, ) Reinforced 939 
Table 37.—Ratio of “e for Different Steel Ratios p............... 940 


Diagram 24.—Values of k for Different Eccentricities, | Tension Side 940 
Diagram 25.—Values of C4, J Reinforced 941 


General Data 


Ty cou. ogy SRS SSIS gail eel ae iowa a 942 
Table 39.—Areas of Groups of Bars of Uniform Size................ 943 
Table 40.—Area of Square Bars in Slabs for Different Spacing....... 944 
Table 41.—Area of Round Bars in Slabs for Different Spacing....... 944 
Slee -ELOMCrUes OL GCCUONS. .. 0. secede ett cee eteeeeree 945 


TABLES IN THE BODY OF THE BOOK 
Following tables are given in the body of the book. 


Properties of Reinforcement 


Areas, Weights and Perimeters of Round and Square Bars.......... 12 
Wire Uecderateitrercementse. MEPS) oS ae’ PUI LS ace, 15 
Properucs a: Mxpanded Metal) 2. oiy en. TE a a. 16 
Properties of Welded Wire Fabric.....:.....-.... 057200000 be. ea ee. 17 
Properties'of Triangle)Mesh Fabricy).. 0...) oo 18 


Bond Stresses 
Length of Lap for Different Stresses in Steel and Diameters of Bars... 414 


Spacing of Bars in a Beam 
Width of Beam Required by Different Number of Round and Square 


ISR oe ale a eee Nn ne OR ee ee 270 
Distribution of Concentrated Load on Wide Slab 
Effective Width of Slab for Concentrated Load.................... 70 
Beam and Slab Design 
Constants in Beam and Slab Design for Selected Unit Stresses. ...... 205 
Distribution of Load on Slab Reinforced in Two Directions......... 212 
Columns 
Stresses Recommended by the Authors for Columns with Longitudinal 
SPST See, IR ae eet Oe ae: SE ae 7 CP eee ete Semen 407 
Stresses Recommended by the Authors for Spiral Columns.......... 422 


Requirements of Building Codes of Various Cities Columns with 
Weoevocimal tteck ONY oes: skeen ew sles 8 a 8 ee ae eae ria ae 409 


878 TABLES AND DIAGRAMS 


Requirements of Boston, Cleveland and Philadelphia Codes for Spiral 

Columns. 2. 0. Prt 
Standard Wire Used for.Spirals. ......--+.+5-> => s)he =o eee 
Recommended Proportion of Live Load on Columns......-.---+---- 


Footing Tables 

Required Depth of Plain Concrete Footing in Terms of Length of Pro- 

FOCtION «esse ene cet ce terete ees 070 0m Muelle ae age cee cae 
Dimensions of Independent Stepped and Sloped Square Footings... a 
Constants C;, for Square Footing in Formulas for Shearing Stresses. . . 
Constants C,, for Rectangular Footing in Formulas for Punching Shear. 
Constants C,, for Square Footing in Formulas for Bending Moments. . 
Constants C,, for Rectangular Footing in Formula for Bending Moment 
Constants C;, for Rectangular Footing in Formula for Diagonal Tension 
Economical Length of Long Side, a, of Rectangular Footing <2. iad 
Constants C,, in Formula for Center of Gravity of Corner Column... . 


Chimney Tables = 


Values of Constants Cp, Cr, z,7 for Different Positions of Neutral Axis. 
Location of Neutral Axis. . 4: 4<- 1...) .4> «+5 gp esis ee ee 


Retaining Wall Tables 


Average Weight of Ordinary Earth before Excavation. .../:1. 9, 22%. 
Values of Constants Cp in Formula for Earth Pressure.........-----. 
Constants Cp in Formula for Length of Base.......-----+++-++++-> 
Spacing of Counterforts............+ +1: Aven Res ote Se Sena 
Thickness of Retaining Wall of Gravity Section.........-+++++++++- 


Miscellaneous Tables 


Weight of Hollow Burned Clay Tile. .......--+++++++seereteee ress 
Weight of Clay Tile Floors.......-----+1++ +++ 55: apse siete ede 
Required Thickness of Brick Wall.......-----+-+-+-++eesssss css 
Dead Loads on Roofs... ee oe oa 
Live Loads Required for Different Buildings..........---. Fe iii nee 

Dead Loads of Floor Finish, Plaster, Partitions, etc........-..-.---- 
Overall Dimensions of Passenger Cars... .....--- ++ se ee eee eters 
Overall Dimensions of Trucks. ......00eeessnenrcneene renter nt os 


PAGE 


423 


TABLES AND DIAGRAMS 879 


Table 1.—Working Unit Stresses. 


Allowable Working 
Stresses. 





Nota-| Percent-| For 








Kind of Stress. tion, |_28¢ of | 2 000 Remarks. 
Crushing} Lb. 
Strength|Concrete, 
at 28 | Lb. per 
Dave. | mq. in. 
(1) (2) (3) Ae (5) 
BCE ed Perea a) City og fs is a 2 a5 t T 
Length of pier 
Exh) COMPLOSSION 6. cca ene ese te 2ar0 450 not to exceed 
4 diameters 
Vertical steel 1 to | Length at a 
Columns (as Mee A ae if 2265 450 pee ys 
described BUS Oe 
on page} Vertical steel 1 to | omen a hl 
561) 4% and spirals d40least 
Oe Sie aa Hah alee 700 P aah gaits 
radius of gy- 
| ration 
moiniaryee st, .. o fe 40 800 Use in beam for- 
Compression mulas 
in extreme } In continuous beams 
Der s. .} adjacent to the| 
BUupporiuss.. > 2). te 45 900 
Shear (punching shear). ..........|./.... 6 120 
Shear (as (Beams without web 
measure of | reinforcement....| v 2 40 
diagonal ( Stresses above 
tension... | Beams with web re- 40 Ib. must be 
| inforcement...... v 6 120 provided for 
with web re- 
inforcement 
Head { Plain: barsiviies. 4: u 4 80 
bap vic a Deformed bars.....} u 5 to 6 |100to120 
( Structural grade....| fs 16 000 |Ib. per 
Steel in ten- sq. | in. 
MOU a fees. Intermediate and 
high carbon steel.| fs 18 000 |Ib. per 


Stja} ins. 
es ee ed 8 ol ee 
* Strengths at 28 days for different proportions of concrete are given on p. 205. See 


Volume II for discussion of strength of concrete. 
+ See Formulas (56), p. 271 and (57), p. 272. 


880 TABLES AND DIAGRAMS 


Table 2.—Constants for Rectangular Beams and Slabs 


To be used in formulas for Depth of Beam, d = oy ord = Na and 


Depth of Slab, d = C1\/M; in formulas for Moment of Resistance, M 


M 
or M = Rbd?; in formula for Area of Reinforcement, A, = pbd, or Aa 


Ratio of Moduli of Steel to Concrete, n 





Working | Working Ratio 
Strength | Strength | Depth of 
of of Neutral 
Steel, |Concrete,} Axis to 
Lb. per | Lb. per | Depth of 

Sq. In. Sq. In. Steel. 








fs te k 
14 000 500 0.349 
550 0.372 
600 0.392 
650 0.409 
700 0.428 
750 0.446 
800 0.462 
850 0.477 
16 000 500 0.319 
550 0.339 
600 0.358 
650 0.378 
700 0.397 
750 0.414 
800 0.429 
850 0.444 
900 0.458 
18 000 500 0.294 
550 0.314 
600 0.333 
650 0.351 
700 0.369 
750 0.385 
800 0.400 
850 0.415 
900 0.429 
20 000 500 0.272 
550 0.292 
600 0.311 
650 0.328 
700 0.344 
750 0.359 
800 0.374 
850 0.389 
900 0.403 











Ratio of 
Moment 
Arm to 
Depth of 


ee 
Lies 


j 





Ratio 
Area of 
Steel to 
Beam 
Above 

Steel. 


0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0. 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 








_ bd? 

jdfs 

= 15: 
Constants. 

For Beams. For 

Slabs 

Cc R C1 
0.114 CEL 0.0329 
0.106 89.4 0.0306 
0.099 102.1 0.0286 
0.093 lib.2 0.0268 
0.088 128.6 0.0254 
0.083 142.2 0.0240 
0.080 156.2 0.0231 
0.077 170.4 0.0222 
0.118 Tio 0.0341 
0.110 83.0 0.0318 
0.103 95.0 0.0297 
0.096 107.5 0.0277 
0.091 120.4 0.0263 
0.086 13320 0.0248 
0.083 146.9 0.0240 
0.079 160.6 0.0228 
0.075 174.5 0.0216 
0.123 66.3 0.0355 
0.113 77.4 0.0326 
0.106 : 88.9 0.0306 
0.099 100.8 0.0286 
0.094 113.1 0.0271 
0.089 126.2 0.0257 
0.085 138.7 0.0245 
0.081 151.9 0.0231 
0.077 165.4 0.0222 
0.127 62.0 0.0367 
0.118 To CG 0.0341 
0.109 83.5 0.0315 
0.103 94.9 0.0297 
0.097 106.7 0.0280 
0.092 118.8 0.0266 
0.087 131.3 0.0251 
0.083 144.0 0.0240 
0.080 157.0 0.0231 








TABLES AND DIAGRAMS 881 
Table 3.—Constants for Rectangular Beams and Slabs 


To be used in formulas for Depth of Beam, d = cy ord. = ee and 











Rb’ 
SE AGE : bd? 
Depth of Slab, d = Ci/M ; in formulas for Moment of Resistance, M = Ca” 
: [ M 
or M = Rbd?; in formula for Area of Reinforcement, A, = pbd, or A, = cafe 
Jas 
Ratio of Moduli of Steel to Concrete, n = 12. 
Working | Working Ratio Ratio of Ratio 
Strength | Strength | Depth of Moment | Area of b Constants. 
of of Neutral | Arm to | Steel to 
Steel, |Concrete,| Axis to epth o Beam 
Lb. per | Lb. per | Depth of Steel Above For Beams. For 
Sq. In. Sq. In. Steel. (1 aa 4 Steel. Slabs. 
3 
fs te k J p Cc R Cy 
14 000 700 0.375 0.875 0.0094 0.093 115.6 0.0268 
750 0.391 0.870 0.0105 0.088 129.2 0.0254 
800 0.407 0.864 0.0116 0.084 141.6 0.0242 
850 0.422 0.860 0.0128 0.081 152.4 0.0234 
16 000 700 0.344 0.885 0.0075 0.097 106.7 0.0280 
750 0.360 0.880 0.0085 0.092 118.8 0.0266 
800 0.375 0.875 0.0094 0.087 i3ivo 0.0251 
850 0.389 0.870 0.0103 0.083 144.0 0.0240 
900 0.403 0.866 0.0113 0.080 157.0 0.0231 
1 000 0.429 0.857 0.0134 0.074 184.0 0.0214 
18 000 700 0.318 0.894 0.0062 0.100 100.0 0.0289 
750 0.333 0.889 0.0070 0.094. kee VL 0.0271 
800 0.348 0.884 0.0077 0.090 123.4 0.0260 
850 0.362 0.879 0.0085 0.086 130.1 0.0248 
900 0.375 0.875 0.0094 0.082 148.8 0.0237 
1 000 0.400 0.867 0.0111 0.076 173.0 0.0219 
20 000 700 0.296 0.901 0.0052 0.103 94.3 0.0297 
750 0.310 0.897 0.0058 0.098 104.1 0.0283 
800 0.324 0.892 0.0065 0.093 115.6 0.0268 
850 0.338 0.887 0.0072 0.088 129.2 0.0254 
900 0.351 0.883 0.0079 0.085 138.3 0.0245 
1 000 0.375 0.875 0.0094 0 1 0.0226 


.078 164. 





Use of Tables 2 and 3.—The constants in Tables 3 and 4 may be used for 
determining concrete dimensions and the required amount of steel for rectangular 
beams and slabs for known bending moment M, and known stresses f., fs, and n. 
Care should be taken to use the table for proper value of n. For 1:2: 4 con- 
crete n = 15 is commonly used. For 1:13:3 concrete n = 12 and1:1:2 
concrete n = 10 are common. 

In formulas for depth, if M is in inch-pounds and } is in inches, the depth will 
be also in inches. With M in foot-pounds, b should be in feet, and d will also be 
in feet. In both cases the constants remain the same. To change bending 
moment from foot-pounds to inch-pounds multiply the foot-pounds by 12. 


770 BRneSaeene ee TEE | 






























































Poo ECE oan SceReeSaeeaEe 
760 
nee 
150 a4 
aon 
Coo 
140 Poth 
rH 
i 
730 HH 
“ 
i 
& 110 
~ 
er a6 Z 
“ PATA TCO 
AA 
~ ay Coageerae 
3 90 Te a A 
> Be enue oo ae ae 2 
7. Pot LA ZL 
80 aia av arenaMe’s 
ava sua8 
Aer 
70 Ad 


ARP a nae 
Zari | vi 

Ai Tt 
Boe 























Ratio of Reinforcement p 


Dracram 1.—Values of R and p for Rectangular Beams and Slabs. 
For Different Values of f,; and fe. 


Use of Diagram 1.—Following problems may be solved by means of this 
diagram: 

1. Determine values of R and the ratio of steel, p, for any given stresses 
f- and fs. 

In solving this problem find point of intersection of the curve corresponding 
to the stress f, with the curve corresponding to the stress f,. Vertical line down 
from this point gives the required ratio 7, while a horizontal line gives the value 
of constant f. 

For intermediate values, interpolate. 

2. Find stresses f,; and f, for known concrete dimensions b and d, known 
area of steel A, and given bending moment M. 


M A Se 
In solving this problem find R from R pies and p from p = 7 Find in 


the diagram the point of intersection of a vertical line corresponding to the 
determined value of p with a horizontal line corresponding to the value of R. 
This point by interpolation between the f, curves gives the stress f, and between 
the f, curves the stress fs. 

882 


TABLES AND DIAGRAMS 883 








Values of jandk 











g wae 





Values of Pp 


DiacraM 2.—Values of J, k, and” for Different Ratios of Steel p. 


c 


Use of Diagram 2.—From this diagram the ratios 7 and k may be obtained 
directly for any ratio of steel p. 


This diagram may also be used for computing stresses in a beam with known 
dimensions b and d known amount of steel As and given bending moment M. 


As : 
In solving this problem compute value of p from p = bd Find from diagram 


the corresponding value of 7 and . Compute stress in steel from formula 
c 


Ss 


ere in which all values are known. Knowing the stress f, and the ratio 
sJ 
3 the stress in concrete may be readily found by dividing the stress by the 
c 
ratio fs 
G 


TABLES AND DIAGRAMS 


884. 





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jO JUSTIOTN| 25, Ee -| qydeq yydeq | xuC ulBveg epIM YU] 9uUC WvIg JOJ JOO IvsUTT dod (MN) pvo'y sary pus pvaq jeg [Voy yqideq 
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[IPI UI GOUT 9uUO SUIBEq JOF SUIPROT OFeG 
sUlveg snonuyuog jo suedg puy—suvog IJe[nsuejoYy— fF 91qu_L, 


885 


TABLES AND DIAGRAMS 


*SaSSol1]S SUIYIOM UIAIS OY} JO} polNbal St yoryM (Judd Jed 1/'0) 2200'0 =@ [2938 JO O1}VI GB IOF SI SIQT, x 
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‘SoYOUL UL YYpIM Aq AT[diy[Nu waq fo yypim fiuv fo poo) afvs og “[ SAINY 




















































































































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0co 9 0900 Go'l GL 2 9°6 SS el” Be NE NE | SSS, 16Sy PS ay SEre (Bhp aS Ge 99ee OR: MCDLAESL (CS ta ede 6 
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sulvag snonujuog jo suvds so110}UJ—sWiveg IepnsuvjyI9Yy—"g a1quy 





TABLES AND DIAGRAMS 


886 




































































8 0 | £0O1| 80 OT|62 0 OI|8sz°0 | #6 |92°0 6 |Z2'0 6 |1Z'0 | #8 |69'0 | #8 |249°0 ) #8 |S9'0 8 |19°0 g |6c'0 | 2 0% 
820 o1lsz'0 | §6 |#2°0 | #6 |€2°0 6 |1Z°0 | #8 |29'0 | #8 |99°0 g |lcg'o | 8g le9'0 | 8 |19'0 | 2 |Z¢'0 | FZ \gg'0 L 61 
921°0 6 |$L°0 6 |69°0 6 |89°0 | #8 |99°0 gs |s9°0 g lzo'0 | $2 |09'0 | #2 |89'0 | FZ |Zg°0 L \e¢‘0 LZ \1g°0 | #9 SI 
02:0 | $8 |29°0 | #8 |99°0 | 8 |£€9°0 g i290 | 2 |s¢°0 | ¥2 |Sg'0 | #2 [990 L \F¢°0 rage Lee L \6r'0 | #9 |9F'0 9 LI 
cg'0 | g iz9'0 | 8 |19'0 | £2 |8g°O | FZ |29°0 L \¥FS'°0 2 jis'0 L \oc‘0 | #9 jog‘o | #9 |ZF'0 | 79 |Sh'0 9 |tF'0 | ¥¢ 9T 
09'0 | 2 |z¢°0 | $2 |¥9°0 | #2 |¥9°0 L les'o | £9 |0¢'o | #9 |Z | #9 |SF°0 tq |pr'0 | 9 |€F'0 9 |2r'0 | $¢ |IF'O ¢ cl 
cc'0 LZ \e¢°0 L |0o¢g"0 L |oc‘o | $9 |9F'0 | #9 |9F'°0 9 |€F'0 9 lero | 9 loro | °9 |Or'0 | FE j28°0 | FS |98°0 G FL 
oc'o | #9 |sr'0 | £9 |¢r'0 | #9 [970 | 9 ler 0] 9 ero | $¢ loro | $¢ [680 | FE |Ze°0 | FE |2Ze°0 ¢ |Fe'0 ¢ jee'0 | fF roa 
9F 0 9 |#F'0 9 |IPF'O 9 |zr'0 | $¢ |sg°0 | FE |6E°0 c |LE8°0 ¢ |¢e'0 ¢ |r 0 c |pe'0 | &% [180 | #F |8z°0 | FF aL 
zero | £¢ |or'o | ¥¢ |ze°0 | Fe jseO| ¢ jse Oo) 9¢°0 | &F lego | &% |ae'0 | FF [TE°O | th |62°0 £p 19Z°0 | F |92°0 F IL 
se'0 ¢ |98°0 ¢ |ge'0 c logo | §F |ze'0 | &F |62°0 | tPF |22°0 £p 11Z'0 | &P |182'0 ¥ |92'0 ¥ |§z°0 F |1z'0 ¥ OL 
120 | ¢ |ze'o | & j0e'0 | fF |82°0 £y loz'0 | §F |22°0 | F |S0°0 7 1t2'0 | F |E2°0 7 |\Fz'0 | $e |zz'0 | #8 |0c'0 | F€ 6 
1z°0 | & 1920 | &F [2z°0 | ¥F |Ss'O| F j€s°0 + lpz'o | €e |zz'o | #e |2a'0 | fe |1Z'0 | FE |61T'O £e Igl‘O | #§ [STO g 8 
ez'0 | ¥F |92'0 | $e |¥2°0 | te |86°0 te loco | €€ [610 | #8 [2410 | #E |9T"O fe 16L°0 € |91°0 | #8 {STO € |rt'o| & i 
eto | €e [gto | ge [z2T°0 | Fe [9T°O | FE [STO € |Z1°0 e |ct°O | § {STO € |FI'0 € |§1'0 © ltt 6 ¢ |O1'0 g 9 
LEG) Fe 19T De 1ST 0 € |F1°0 ele o € |IT’0 € |IT°0 ¢ O10 | € (0OL'0 € |60'0 € |80°0 ¢ |20°0 g c 
II°0 | #& 1010 | #8 60°09 € |60°0 € |80°0 ¢ 120°0 | ¢€ 120°0 | §€ |20°0 € |90°0 ¢ |90'0 | & |¢0'°0 € |t0°0 e 7 
ate S ice ul a]. |; i eae 5 ee ay are Le ed ai Pee ale Se lee ial te ‘ul P 
sy | ¥ sy | ¥ sh oom ae | ey iy ep tee | ey eee ty. OL ca ge Bay ad oa a) aod eek Oe ea SS Bed 4 Y 

* OST x OFT PITAL * SIT « OOT x 06 x 08 iL « OL * 09 « OS « OP uads 

4% (00,7 e1enbg rod spunog poy 9AIT 
set Nal <A eee ae ee ea me 
"Y}pIM Jo 4Oo} Jod [90}8 JO UOTJDVS SSOID JO BOTY = ey L-Y¥u=P ‘qydop oATpoeHA “qvBIS Jo ssouyory} [BIOL = Y 
‘Jaaq8 JO Bole puB qus Jo [dep OF %0z ppe (suvds pezzoddns) 52h W 0g 
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v4 


a 
pvoT OATT UdATY) OF pormnbey JUDUIII1O FUTOY pus qeis jo Sssouxory, 


‘SqBIS snonuyu0D—'9 21qeL 


887 


TABLES AND DIAGRAMS 


*PBoy aAr 
porepisuos Ott SENT 

I qou 71) orqyt PVOT IOSIVT ° 

1) OIG}POURIS IOJ MOT[S qj eT 0} Surpuodse 

s : I 1109 
Iyytjouwss pesoduiedns 10; I[¥ Joputo WyIM Bumo0op poom sO ( 
> jou ynq quis 8 : IO (QBs [ein 
4a10U0dD JO JYBIOM I 4oni4s Jo y1vd s¥ 
: O} opeuUl SI soueM 
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Oi tT €1/90° 
590 °T © ; 
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60 | ftrlre: z1/96°0 | ¢ : 6°0 | 2126" 
ae £111/76°0 | £111 z6° eTTI¢6°0 | 2 : L6°0 | ,¢11/6" 
: 88°00 | $01 : 0 Il|06'0 06°0 I 28° -L1|68'0 
T8°0 orl6z: £9Tle8°0 £0o1/F8'0 | ZoTIEs° 201/28 0 
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92°0 £6 ‘ 0 76/2550 z OT/82°0 OL : 0 OL|IO8 °O zZ 
ca erg RG ae rL°0 | $6 |¢2" 6 je20 1.4 12°0 | #6 |¢z'0 | § 26 |21°0 | $6 
69°0O 6 ‘ g L°0 6 lez" 16 |7L°O 6 z 0 | £6 |*2°0 T SI 
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LF°0 | £9 |9F° oo | £ lze01 L |e¢°0 O14} .£58S"0 tL v1 
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TABLES AND DIAGRAMS 


888 

















fens (ue tae sim 























ther et ane 





882 | IPE | GOP | 18h | G29 | 069 

peek porte Tee Te eee oe ee ee ae ees 
coe PRGR UGE | PRE LOR | SOG Co eee teed ieee eae ee 
G6ial OSE TERE LOPE ROLY OGD Oe | OR eee ee 
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cpt | cat | ziz | sez | ple | ese | BAe We eA ee Beer I me mci te ee eg © 
Det COE Le SERTS Zee” PILE SE LOTR eh eee een eee 
Oh UL SELShGeTe MRT: | EGP S82. | GRC OEP rc a) ke ok 2 ee 
ty eG GCE LOOT | POL POPE | GOR UISLE (GLE oe ce ee 
6 REGO | tel | BSL | 00S ce Ole. c0r Ole] ae ioe: a ee ae 
ep 6G PRL EIOT | GSP 1 SUL) GOS 0G.) Whe 1 cer | eced ele oe ia 
6g | ep ae 81. en 4 Corer Ter BST otc hte Woe | SEP pre ey 
OL hue reo" | WaerGe PT OCrSITZGIT eietase | Ske | rose | 
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gt | 62 | F< /-8O | 88 | ZI | OAT | 68% | SPE | Tes 

Zb | $% 1°28 |9S' 1.08 | SIL | 99L | Sra | ee 

toa eban | £ve bald t SOLO Tere 

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‘sqe[g peyroddng Ajdung—) s1quy 


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= jy ‘juewou suIpusg 


889 


TABLES AND DIAGRAMS 











868 | FOF | FS | GEO 
ce | OIF | 88F | GZg 
cTe | OLE | Ser | FIS 
81Z | 268 | 988 | SSF 
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19% | Sze | 668 
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TABLES AND DIAGRAMS 


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TABLES AND DIAGRAMS 









































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894 TABLES AND DIAGRAMS 


DIAGRAMS FOR T-BEAMS. 







































































































































































































































































OL rat 1 0.015 

S & 
x 
Hoes 0.010 
© g 
= 2 
S 0.005 0.0055 

0 
0 0.05 OI02 S078 S 0. <5 dbs 
Values of 4 
i wih 
DracraM 3.—Values of Maximum Steel Ratio pm for Different Ratios 7 
Governed by Compression Stresses. (See p. 216.) 
1000 & 
900 
=, u 
800 se 

: a . Se 

SS) 

ae 702 Face BEES == Seeee o 

pie dene : SSenseecses > 

% os cS =e a = = 3 

3 600 = = % 

s : 8 
= 







































































UTA TT 


aan 
HAN ANN 











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tt LH 














a “alo 015 oS "930 035 
Values of # 


For Example, see p. 896. 


t , 
Dracram 4.—Values of Constants, C, for Different Ratios 7 In Formula for 


M 
Minimum Depth, d = ~~. (Seep. 216.) 
Cqbt 


895 


TABLES AND DIAGRAMS 


DIAGRAMS FOR T-BEAMS 


“qd fo sanjo, 


0.35 























0.30 








025 











Ge Sata Taal 


020 























O15 
Values of 









































AWE 
BesA\\ge 


010 














maar VA 
@ YT AW 
be | | IAN 





0.05 























HEE 


“4 fo sanjo/ 





t 
- 


DiaGcram 5.—Values of Maximum Steel Ratio p», for Different Ratios 


(See p. 216.) 


Governed by Compression Stresses. 




















P5-- fo sano, 
S S s) S S Ss) S S 
S S S S Ss S S Ss 
iS D a ~ © % -s “o 
i | iH 
| 
| 
| 
| 
| 4 
TT i / 
LY {|| 1 
LTT VALLETTA 
LETTE TTT AT TTT BA AT 
PERERORURDGRD GEDA GR ORER) CRROG) ARRGR) /URRRORY 400) NED AUUED 
RR eT Ra Re ASURURRUEARD? AURLY-ANRUAY UOLARAY ABRY (ORY (ALA 
HTT A 
LEECH OEE TTT UE EEE IL 
HTT NATTA HH 
HEHE AULT QO META 
HTT ATT TTT Se 9 AL anil il 
TAT HTN ATT 9 PA I 
CTT ATTTTT TPT ATT SATA 
PORREDY CCUURRRURRGL NY ZORUERD) (01 SAT MYT 
A AUT ATT ASA SAO FLT TTT 
a HELL ATTA LOA OAT] 
ATTA 
TTT TT HALT ATT TTT TT 
HTT ATA 
HTT AT 
HECELCLL AA TEECt REHEAT HERE 
TTA 
HITT TTA 
HATTA 
ITAA 
nnd 19% LATTA LT 
Vaal 41 AUT | 
S S S S S S S 
Ss S S SS S S 
D 9 ~~ ey Ly vt ~ 






























































































































































Pp fo sanjno, 















































































































0.20 025 


Values of 4 


For Example, see p. 896. 


In Formula for 


t 
7 


? 


Diacram 6.—Values of Constants, Ca, for Different Ratios 


(See p. 216.) 


Hake 
Cor 


Minimum Depth, d 


896 


TABLES AND DIAGRAMS 


Table 13.—Use to Find Stresses f, and f, in T-Beams. 


(See p. 898.) 


As . 
Values of k, 7, and C'r, for Different Values of a (See pp. 224 and 898.) 





Ratio 
nAs 
bt 


it 


ooo Coco coco cooe oooo oooo 
wo 
bo 


Ratio 
of 


Moduli 


n 





Value of k. 


Use to find 7 and CT below. 





« 0.10 


. 136 
.208 
. 269 


.321 
.345 
. 367 


387 
406 
.424 


441 
457 
472 


(aayes) “CeKola) (aioleny haleayer 


Me) A 
On 
) 











Ratios of Thickness of Flange to Depth of Beam, 4 





O12 420514 SOM GeO as 
0.145} 0.154] 0.164 
0:217| 0.225) 0.233] 0.242 
0.277| 0.285] 0.292} 0.300 
0.329] 0.336] 0.343] 0.350 
0.352] 0.359] 0.365} 0.372 
0.373] 0.3880] 0.387] 0.393 
0.393] 0.400} 0.406) 0.413 
0.412| 0.419] 0.425) 0.431 
0.430} 0.436] 0.442] 0.448 
0.447) 0.453} 0.459] 0.465 
0.463] 0.469) 0.474] 0.480 
0.478; 0.483) 0.489] 0.494 
Values of 7. 





0.20 


. 250 
. 308 


357 
379 
.400 


.419 
.437 
.455 


.471 
.486 
. 500 


ooo ,ocoo ooo oo 


0.22 


. 258 
.315 


. 364 
386 
407 


.426 
.444 
.461 


.476 
.491 
. 006 


coo _ Sooo coco oo 


0.24 |0.26 | 0.28 | 0.30 | 0.32 


. 26710. 
.323)0. 


.37110. 
.393)0. 
.413]0. 


.432)0. 
.450]0. 
.467/0. 


.482/0. 
.497)0. 
.511/0. 


ooo ooo, occ oo 


Use to find fs in Formula fs = 


275)0.283 








331/0.338]0.346]0 
379|0.386]0.393 0. 
400|0.407 |0.414/0 
420|0 .427 |0.433)0. 
439|0.445/0.452|0 
456|0.462/0.469|0 
473/0.479/0.485)0. 
488|0.494/0.500/0 
503]0.509|0.514/0. 
517/0.522)0.528)0. 
M 

Asjd’ 


Ratios of Thickness of Flange to Depth of Beam, 4 























sake 

.914)0. 
.912)0. 
.910)0. 


.908)0. 
.907)0. 
.905/0. 
.904/0. 


.903)0. 
.902/0. 
.901)0. 
.900)0. 


.899)0. 
.899/0. 
250410. 
.897)0. 


.896/0. 
.896)0. 
.895)0. 


oEeo coco ,oCcoo coco CoCo 


(See p. 


304 


400 


421 


440 


.458 
ATS 


491 


. 506 


520 
533 


—— 


913 
909 
906/0.907 


904/0.904 
902/0.901|0.900 
900|0 .899 |0.897 


8938/0. 896 |0.894/0. 


897/0.895/0.892/0. 
896|0.893|0.890)0. 
894|0.891/0.888)0. 
893|0.890/0. 8860. 


892/0.889/0.884)0. 
891|0.887 /0.883)0. 
890|0.885|0.880)0. 
889 |0.883|0.880)|0. 


888/0.881|0.879|0. 
887|0.879|0.878)0. 
886/0.877 |0.870)0. 


224). 


0.20 | 0.22 | 0.24 | 0.26 | 0.28 | 0.380 | 0.32 





0.22 


0.028) 0.032) 0.035 


OPTS Fs 20RISS OSLG Owls 
0.948) 0.940) 0.937) 0.933) 0.930 
0.948; 0.940} 0.935] 0.931] 0.927 
0.947} 0.939| 0.934} 0.929) 0.925) 0.924 
0.947] 0.938) 0.933} 0.928] 0.923) 0.921 
0.946! 0.938] 0.933] 0.927} 0.922] 0.919 
0.946] 0.937) 0.932} 0.926} 0.920) 0.917 
0.946] 0.937) 0.931} 0.925) 0.919] 0.916 
0.946] 0.936] 0.931) 0.924) 0.918) 0.914 
0.945) 0.936] 0.930! 0.924} 0.917) 0.913 
0.945} 0.936] 0.930} 0.923) 0.916) 0.912 
0.945] 0.935| 0.929) 0.922) 0.915) 0.911 
0.945} 0.935) 0.929} 0.922) 0.915) 0.910 
0.945) 0.935] 0:929|} 0.921] 0.914] 0.909 
0.944} 0.935) 0.928) 0.921} 0.913) 0.908 
0.944] 0.934] 0.928) 0.920} 0.913| 0.907 
0.944] 0.934} 0.928} 0.920) 0.912) 0.907 
0.944} 0.934] 0.927) 0.920} 0.912) 0.906 
0.944) 0.934] 0.927) 0.919} 0.911| 0.905 
0.944] 0.933; 0.927} 0.919} 0.911} 0.904 
0.943} 0.933!) 0.926} 0.918) 0.910} 0.903 
0.943! 0.932) 0.926) 0.918) 0.910) 0.903 
0.943} 0.932) 0.925}) 0.917} 0.909) 0.902 
0.943) 0.931) 0.925! 0.916!) 0.909 on 
Values of CT in Formula fe = CTrfs. 
Values of & 
0.24 0.26 0.28 0.30 0.32 0.34 


0.36 


O38 (P0540 e0e 


42 


0.019} 0.021} 0.023) 0 026) 0.029) 0.031) 0.034] 0.038] 0.041] 0.044] 0. 
0.024} 0.026; 0.029] 0.032) 0.036) 0.039} 0.043} 0.047| 0.051] 0.055] O. 
| 0.039| 0.043) 0.047 OO 0.056} 0.061} 0.067] 0. 


048 
060 
073 


x 


or 


TABLES AND DIAGRAMS 897 


USE OF DIAGRAMS AND TABLES FOR T-BEAMS 


Use of Diagrams 3 and 5.—These diagrams give maximum ratio of steel, 
Pm, Which can be used in a T-beam without exceeding the allowable compression 
stresses in concrete. Also they may be used to check a design for compression 
stresses. 


Example.—Given M = 1800000 in. -lb., f. = 650, fs = 16000, n = = 15, 
d=18in.,?=4im.,6=58 in. Find Peneired area of steel and see hee: 
gies stresses are satisfactory. 


M 
Solution.—Area of steel is found from A, = fja where 7 is assumed as 0.89. 
sJ 
1 800 000 7202 


Aes MM alence dhe tation of ateciiia pce 
*~ 16 000 x 0.89 X 18 OE TS AR ne Be ea a eR 








t 4 
0.0067. For Car 0.222, and the stresses, f, =650, f; = 16000 from 


Diagram 3 the ratio is pm = 0.0053. This is larger than the computed value 
of p, consequently the compression stresses are larger than the allowable. 
By locating p = 0.0067 in the same ae it is found that the compres- 
sion stresses are about 770 lb. per sq. in. Therefore the depth, d = 18 in. 
cannot be used unless the compression falas is made thicker or compression 
steel is used. Ordinarily it is cheapest to increase the depth of beam. 


Use of Diagrams 4 and 6.—These diagrams give the minimum depth of 
T-Beam as governed by compression stresses for any given dimensions b and t, 
given bending moment M and specified unit stresses f, and fe. 

Example.—Find minimum depth of T-beam as governed by compression 
stresses fe, when M = 1 500 000 in. lb., t = 4 in., b = 58 in., and the stresses 
are fc = 650, fs = 16 000, and n = 15. 


M 
Solution.—Minimum depth will be found from formula d = Gwe in which 
d 


all values except Cg are known. Assume : = 0.24. (This value may cor- 

respond to minimum depth required by shearing stresses.) From the diagram 4, 

corresponding to specified stress f,, and assumed “ the constant is Cg = 386. 
Minimum depth therefore is 


1 500 000 


= ————__ = 16.7 in. 
B86 X58 X40 


898 TABLES AND DIAGRAMS 


Use of Table 13.—Rule for use of Table 13: Find k, use k to determine 7 and 





thus f; in formula A, = Use k also to find Cr and hence fe. 


jaf s 
Example.—Given M = 1 900 000 in. Ib. resisted by a T-beam with the fol- 
lowing dimensions: 6 = 58 in., t = 4 in., d = 22 in. A, = 6.2 sq. in. Find 
stresses f, and fs, assuming m = 15. 
nAy - 15°%-6 
bt §=58X4 
Find in Table 13, corresponding to the above ratios, k = 0.35. With this value 


t 4 
Solution.—Compute the ratios = 0.4 and 7 ap =. 0.18. 
t : 
of k and on 0.18, find from the second part of the table 7 = 0.920, and, by inter- 


1 900 000 
polation, Cr = 0.036. Stresses in steel therefore fs = 0.920 X 22 X 6.2 Fe 


15 200 lb. per sq. in., and f, = Crfs = 0.036 X 15 200 = 547 Ib. per sq. in. 


BASIS FOR TABLE 14 


Asf: . 4 
J , in which 
010 





Table 14 is based on the general formulas = 


s = spacing of stirrups, in inches; 
A, = area of two legs of stirrups, in square inches} 
f, = tension in stirrups = 16 000 lb. per sq. in.; 
v, = unit shear at the point under consideration to be resisted by stirrups, 
pounds per square inch. 
V = average external shear at the point under consideration in pounds; 
»y’ = unit shear to be resisted by concrete, pounds per square inch. 


b = width of rectangular beam or width of stem of T-Beam in inches. 


I 


The value », depends upon the method of distribution of diagonal tension 
between the stirrups and the concrete. 
For Method 1, where concrete is assumed to resist a definite amount of 
’ Vv : 
diayonal tension, 1. =7—> —v’. For this the above formula for spacing 


bid 
changes to s = a which is identical with Formula (40), p. 249. 
a —v'b 
For Method 2, where concrete is assumed to resist one-third of diagonal 
tension and the stirrup the balance, 01 = > ae This substituted in the above 


3 jd 
; . 1.5Asfs ° . . . ® 
eoneral formula gives s = v which is identical with Formula (44). 





jd 
Values of v: may be taken from a shearing stress diagram prepared as 
explained on p. 252, 


TABLES AND DIAGRAMS 899 


Table 14.—Spacing of Stirrups. (See p. 902.) 
























































Spacing of Stirrups, in Inches 

Unit Shear 

to be Resisted Width of Beam }, or Stem b’, in Inches 

by Stirrups, | . 

Je a AUR aaa haa | 8 | 10 | 12 | 14 | 16 | is | 20 | 22 24 

\ | 
+ in. rd. U-Stirrup 
10 26 20 16 j 13 11 10a) 59 8 fe 6 
20 Loo LU 8 6 5) 5) 4 4 3 3 
30 9 6 9) a uy 3 3 2 2 2 
40 6 +) 4 3 3 2 2 2 2 2 
50 > 4 3 2 2 2 Pl ae 
60 4 3 3 2 2 2 
80 3 2 2 2 
100 3 2 
120 2 2 
140 2 | 
% in. rd. U-Stirrup 

10 59 44 35 29 25 22 20 18 lor 15 
20 29 22 18 15 13 Lies 10 9 8 t 
30 19 15 12 10 8 7 6 6 5) 5 
40 15 11 9 ie 6 5 5 4 4 4 
50 11 8 fi 5 5 4 4 3 3 3 
60 10 r) 6 5) 4 4 3 3 3 2 
80 7 o 4 4 3 3 2 2 2 Z 
100 6 4 3 3 2 2 2 2 
120 5 4 3 2 2 2 2 
140 4 3 2 2 2 

















SS a 


3 in. rd. U-Stirrup 




















10 Peete 00-1 ba 4G? 1 BO Pra) SI 28 | 26 
20 B21 7-00 fh iol 2G. dees 0 hay 16 14 | 13 
30 BO 26 Pee 21 17 15 13 11 10 9 9 
40 26 | 20 16 13 11 10 2 8 (s 6 
50 20 15 12 10 8 és 6 6 5 5 
60 Li 13 10 S 7 6 6 5 5) 4 
80 13 10 8 6 5 5 4 4 3 3 
100 10 8 6 5) 4 4 3 3 3 

120 9 6 5 4 4 3 3 3 

140 7 5 t 4 3 3 








10 1338 |100 | 80 | 67 | 57 | 50 | 44 | 40 | 36 | 33 
20 Gree) tO 33: We Sy t 25° 1225). 201g 17 
30 44. | 33 | 26 | 22 19 16 15 13 12 11 
40 a Pa ee A ae 14 12 11 10 9 8 
50 25 19 15 12 11 9 8 7 7 6 
60 22 if 13 1B! 9 8 7 7 6 5 
80 17 12 10 8 7 6 5) 5 4 4 
100 13 10 8 7 6 5 4 4 4 3 
120 it 8 fs 5 5 4 4 3 3 3 
140 9 ri 6 5 4 3 3 3 





For explanation, see pp. 252 and 898. For example see p. 902. 


900 TABLES AND DIAGRAMS 


Table 15.—Number of U-Stirrups in Uniformly Loaded Beam. 
(See p. 902.) 


Number of Stirrups per Beam is 2Ns = Got 


2N, = number of stirrups per entire beam; 
1 = span of beam in feet; 
b = breadth of beam in inches (in T-beam, breadth of stem); 
v = shearing unit stress in beam at support in pounds per square inch; 


~ 


»’ <allowable shearing unit stress (or diagonal tension) in concrete alone in 
pounds per square inch; 


C,, = constant from table below. 


Values of Constant Cn for Finding Number of Stirrups in Beam 


I 














Shearing 1-In. 3-In. z5-In. 3-In. 3-In. 
Unit Stress Round Round Round Round Round 
at Support. | U-Stirrup. U-Stirrup. | U-Stirrup. | U-Stirrup. U-Stirrup. 

v A, =,.008.| Ay = 222 Adee A; =. 39 | As = .61 
60 0.017 0.011 0.008 0.006 0.004 
65 0.024 0,016... 2)., 0.08 0.009 0.006 
70 0.032 0.022 0.016 0.012 0.008 
75 0.041 0.028 0.020 0.016 0.010 
80 0.050 0.034 0.025 0.019 0.012 
85 0.059 0.041 0.030 0.023 - 0.015 
90 0.069 0.047 0.035 0.027 0.017 
95 0.080 0.054 0.040 0.031 0.020 
100 0.090 0.061 0.045 0.035 0.022 
105 0.100 0.068 0.050 0.039 0.025 
110 O,111 0.076 0.056 0.0438 0.027 
115 Onl 22 0.083 0.061 0.047 0.030 
120 0.1383 0.091 0.067 0.051 0.033 
130 0.156 0.106 0.078 0.060 0.038 
140 0.178 0.121 0.089 0.068 0.044 — 
150 0.202 0.187 0.101 0.077 0.050 
160 Op220 0.153 0.112 0.086 0.055 
170 0.248 0.169 0.124 0.095 0.061 
180 0:272 0.185 0.136 0.104 0.067 
Pe 
: 6(v — 0’)? , ‘ : 
Nortr.—Table is based on formula 2N'; = | Aes in which Ag is area of 


two legs of the U-stirrup. v’ = 40 lb. per sq. in., and fs = 16 000 lb. per sq. in. 

Method 1 of distribution of diagonal tension between concrete and stirrups 
is used. Concrete assumed to resist 40 pounds per square inch and stirrups » 
the balance. (See p. 246.) 


TABLES AND DIAGRAMS 901 


Table 16.—Spacing of Stirrups in Beams with Uniformly Distributed Loading. 
? (See p. 902.) 
Spacing in Inches, s = Cyl 

1 = span of beam in feet; 
N, = number of stirrups in each end of beam; 

v = shearing unit stress in beam at support in pounds per square inch; 

vy’ = allowable shearing unit stress (or diagonal tension) in concrete alone in 

pounds per square inch. 





Values of Constant C» for Finding Spacing of Stirrups 






































































































































































































































v = 60 7 = 70 
Ng _| 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th | 9th |10th}| 1st | 2nd ard | 4th | 5th | 6th | 7th | 8th | 9th [10th 
2 |0.29/1.00 0.38/1.29 
3 |0.18|0.39|0.78 0.24/0.50/1.01 
4 |0.13/0.29/0.36\0.71 0.17|0.37/0.47|0.91 
5 (0.10/0.22|0.26/0.32/0. 64 0.13/0.2910.3310.42|0.82 
6 (0.09|0.18,0.2010.24/0.30|0.58 0.11/0.23|0.26|0.31|0.39/0.74 
7 |0.07/0.15/0.17|0.19/0.22)0.28|0.54 0.090. 1910.22|0.25]0.28/0.36/0.69 
8 10.0610. 13/0. 14/0. 16/0. 18/0.210.26/0.50). 0.08|0.17/0. 19|0.21|0.23/0.27/0.33|0. 64 
9 |0.05|0.11/0. 1310. 14/0. 15,0. 17|0.20|0.2510.47 0.0710. 15/0. 16/0. 18|0.19|0.21/0.25|0.32|0. 61 
10 |0.030.1010.11/0 12/0. 13/0. 14/0. 16/0.19|0.23/0.45||0.06|0. 13/0. 15/0. 16|0.17/0.18|0.20/0.24,0.30.0.58 
» = 80 v = 90 
N, | tst | 2nd) 3rd | 4th | 5th | 6th | 7th | sth | 9th |10th|/ 1st | 2nd) 3rd 4th | 5th | 6th | 7th | 8th | 9th |10th 
2 |0.44/1.50 '0.49|1.68 
3 (0.23/0.58|1.18 0.31/0.65)1.32 
4 |0.20\0.4310.55/1.06 0.2210.4910.61/1.19 
5 |0.16/0.34,0.39/0.49|0.95 0. 17|0.38]/0.44|0.55)1.07 
6 |0.13|0.27/0.31|0.36|0.45|0.86 0. 1510.31/0.34|0.400.50/0.97 
7 |0.11|0.22|0.25(0.29/0.33|0.42|0.81 0. 12/0.25/0.29/0.32|0.37/0.47|0.90 
8 |0.09'0.20/0.22/0.24|0.27/0.31|0.3810.75 0.11/0.22|0.24/0.2710.30/0.35|0.43|0.84 
9 |0.03\0.17/0. 190. 21|0.23/0.25|0.30|0.37|0. 70 0.090. 19|0.21|0.23|0. 25/0. 28|0.33/0.42|0.79 
10 |0.07/0.16/0.17\0.1810.19/0.2110.2410.280.35/0.67||0.08/0. 17/0. 19/0. 20/0. 22/0. 2310.26 0.31/0.3910 76 
» = 100 » = 120 
Nz | ast | 2nd| 3rd|4th| sth| 6th | 7th | sth | 9th |10th|| 1st | 2nd] 3rd | 4th | 5th | 6th | 7th | 8th | 9th | 10th 
2 |0.53/1.80] 0.58|2.00 
3 |0.33/0.70|1.41 0.37|0.78)1.57 
4 |0.24/0.5210.66)1.27 0.26/0.58|0.73/1.41 
5 |0.19/0.40|0.47|0.58|1.14 0.210.45|0.52/0.65|1.27 
6 |0.16(0.33|0.37)0.43 0.54/1.04 0. 18/0.36|0.41/0.48|0.60/1.15 
7 |0.13/0.27/0.31/0.35|0.40/0.50|0.97 0. 14/0. 30/0.34|0.39|0.44/0.56/1.07 
8 |0.11|0.240.30,0.29:0 32.0.37/0.46/0.90| 0. 13/0.26/0.29|0.32/0.36/0.41|0.51|1.00 
9 |0.1010.21/0.23)0 250 27,0.30/0.36)0.45/0.84 0. 11/0.23/0.25/0.2810.30|0.33/0.40/0.50|0.94 
10 [0.0910 190.2016. 22/0. 2310. 2510.28/0.33'0.4210.81/|0. 10/0. 21/0.23|0. 24/0. 260.2810. 310.37|0.47/0.90 
y = 140 » = 160 
Nz | 1st | 2nd| 3rd | 4th | 5th | 6th | 7th | sth | 9th {10th)) 1st | 2nd| 3rd | 4th | 5th | 6th 7th | 8th | 9th |10th 
2 |0.63/2.15 0,66/2..25 
3 |0.4010.84/1. 68 0.41/0.88|1.76 
4 |0.2810.62/0.78|1.52 0. 30/0.65/0.82|1.59 
5 |0.22|0.4810.56|0.70|1.37 0.2310.51/0.58|0.73|1.43 
6 |0.19/0.39'0 44.0.52|0.65|1.24 0.2010.41.0.46|0.54/0.68/1.30 
7 |0.15|0 32|0.36,0.41|0.47/0.60/1.15 0 16/0.34|0.38|0.43|0.50/0.63|1.21 
8 |0.13|0.28/0.310.35|0.39|0.44/0.55|1.07 0. 14/0.30/0.32|0.36|0.40/0.47|0.58|1.12 
9 |0.12|0.25:0.2710.30|0.32|0.3610.42/0.53/1.01 0. 12/0 260.29|0.31|0.34/0.37|0.40|0.56/1.06 
10 10.1110 22/0 2410 260.2810. 3010.3410.4010.50/0.971/0. 1110. 23/0. 2510.2710. 2910. 3110. 3510. 4210.52/1.01 












































If larger number of stirrups are used divide the number by 2, find the spacing 
for this number from the table, and place intermediate stirrups between. 


902 TABLES AND DIAGRAMS 


USE OF TABLES FOR SPACING OF STIRRUPS. 


Table 14.—This table may be used for any type of loading. Proceed as 
follows: 

Compute shearing unit stress, v, at the supports and, in case of concentrated 
loads, at the points of concentration. 

Draw shear diagram. Free hand drawing on cross section paper is sufficient. 

Mark off shear resisted by concrete. 

Select diameter of stirrup. 

Scale at any desired point the shearing stress to be resisted by stirrups. In 
the table, opposite this stress, in section for the selected diameter of stirrup and 
the proper width of beam, find the required spacing of stirrup. 

This procedure may be repeated until all stirrups are located. Often each 
spacing is repeated several times so that only three or four values need to be 
found from table. 

When the spacing of stirrups found from the table is larger than the limiting 
spacing as given on p. 250, it should be made equal to the limiting spacing. This 
reduced spacing should be repeated until all diagonal tension is taken care of. 

Table 15.—This table gives number of stirrups for beams with uniform load- 
ing only. Proceed as follows: 

Select. diameter of stirrup. Find unit shear at support, v. Find constant 
C,, opposite this stress in section corresponding to the selected diameter of stirrup. 
This constant multiplied by net span / in feet and breadth of stem 6° in inches 
gives required number of stirrups. The spacing of stirrups are found as given 
below. 

If the number of stirrups from the table is used for estimating of the amount 
of steel, it should be increased by 15 to 20 per cent to take care of the additional 
stirrups required where the theoretical spacing exceeds the limiting spacing as 
given on p. 250. However, to obtain the spacing by Table 16 use the com- 
puted number of stirrups (and not the increased number). 

Table 16.—After computing the number of stirrups 2. NV, as explained above, 
find their spacing as follows: 

Divide total number of stirrups by 2 to get Ns, the number of stirrups at one 
end. 

From the section of the table nearest to the shearing unit stress at the support 
v, take the line of dimensions corresponding to the number of stirrups at one end, 
N,. These multiplied by the net span, in feet, give the spacing of stirrups. 
Mark the spacing on the beam. If any spacing exceeds the limiting spacing as 
given on page 250, it should be reduced. Sufficient number of stirrups must then 
be added to the computed value of 2N, to cover the whole section requiring 
stirrups. 


TABLES AND DIAGRAMS 903 


DESIGN OF BEAMS WITH COMPRESSION STEEL. 


Diagrams 7 to 10.—For the design of beams with steel in top and bottom 
Diagrams 7 to 10, pp. 904 to 907, may be used. Diagrams 11 and 12, pp. 908 
and 909, may be used also but are less convenient except for stresses not covered 
in the others. The use of the diagrams is illustrated as follows: 


Example.—Given, bending moment, M = 2000000; available depth and 
breadth of beam, 32 in. and 14 in.; allowable stresses fs = 18 000, and f, = 750; 
and n = 15. The depth of the compression steel is ad = 2 in. 


Determine the amount of tensile and compressive steel. 


2 
poluion. since A = 32 in., d = 29 in., and a = noes 0.069. From the 


uM : 
formula A, = —-, where we may assume Jj =0.9 (see p. 240), we have 


jaf s 
A ayo 4.3 12 &G t As oe 0.0106. 
= ——______—__ = 4.3 gq. in. ompute p = — = ——— = 
=~ 0.9 x 29 X 18000 se Ns tonite Thalia nag: 


Refer to Diagram 9 in the section for f; = 18 000 and f, = 750, and find the value 
of p’ = 0.0052 corresponding to a = 0.06 and p; = 0.0106. 

Check the value of 7 by referring to Diagram 18, p. 910, and recompute if 
necessary. 

Further explanation of use of diagrams are given on pp. 287 to 240. 


REVIEW OF BEAMS WITH COMPRESSION STEEL. 


Diagrams 11 and 12.—For review of beams with steel in top and bottom, 
where it is required to determine the stresses when the dimensions of the beam 
and steel area are given, Diagrams 9 and 10, pp. 906 and 907, are to be used, as 
illustrated below. 


Example.—Given A, = 3.5sq.in.; M = 1 230000 in. lb.; A,’ = 2.0 sq. in.; 
and. ratio z = 1.75. The depth of compressive steel is ad = 1.5 in. Find 
fs and fe. ‘ 

Solution.—Since h = 24.5 in., d = 23 in. and a = a = 0.065. Compute 


5 2.0 
p= a OF0152, and.’ =. -—— = 0.0087, From Diagram 8 
23 x ifm 235 x 10 


(p. 905) fora = 0.06 and p; = 0.0152 and p’ = 0.0087 we have, by interpolation, 
where j = 0.9 (see p. 240) 











rs = 1.6. From the formula f, = Seis 


1 230 000 ; 
ee 2 17 200: Ib per. sd: Inz.-.mmce 
ee 2 *.= p19) i235 3.8 Eee 15f. 


17 200 : 
and f; = 17 200 we have, f, = (anes = 880 lb. per sq. in. 


Check the value of 7 by referring to Diagram 13, p. 910, and recompute if 
necessary. 
Further explanation of use of diagrams are given on pp. 237 to 240. 


a value of 





108, 





















































































































































































































































































004 
= 003 
/. [ae a 
: ey 
x aes 
= 02 +S 
aes Se 
N eas Roo (ai Ea — 
==== Ql 
rF—_ 0 
Values of p, 
Ff. = 650, f, = 16,000,n =15 F750» fq = 16 000,n = 18 
aod 0.030 et ee 
Ms WR BE 
BSI RE KORE 
BE MER che sek 
0025 
025 Te LALA 
EEe I I ee eee 
003 nd EG 
YA ee Reel es IT ee Nee 
aA oon (ee ee gL 
eae 020 | isa eee 
KR Bees 
~~ mer 
2 002 —— 0015 
aS 
Ms a aa 
a3) Raasryee, 
8 [ta 0010 
> wi : eee 





VLLNLL_| 
at | 


NI 
S 
N 


007 














Values 


of P, 
F800, f,= 16000, n=I5 


9, = ratio of tension steel; iy 











NS SS IN So eS 
= 5 5 5-8) 8 Sesvare 
so:.S SS Ss 3S 3°95 0 
Values of Pp, 
fF. = 900, 752 16.000, a= 15 





ratio of compression steel; a = ratio of 


depth of compression steel to depth of tension steel. 


Diacram 7.—Ratio of Compression Steel, p’ for Given Ratio of Tension Steel p1. 
(See p. 903.) 


Stresses in Concrete, fe 
Stress in Steel, ts 
and n 


650, 750, 800, and 900; 
16 000; 

15. 

904 


I 


006 005 
































0.05 








a 
| 
1 


004 


tH 

PN 
NV] 
bee 
nl 
ITN 


































004 








Hill 
Hint 
UT 
TT 
ane 

ni 

I 
nt 

















of p! 














i 
‘ 
: 
N 


Unt 
TH 
a 
a 
NS 
iN 





003 








Ul 
Sau 
N 
N 
N 
! 








NT] 
ut 4 
NN 
N 
‘ 
: 





[ 9 STC LY LR LF NCD CE 





A 
uy 
i 





N 





Values 
Ny 
WO 
II 


we 4 
fee Ue Ls 
LP CL | 


002 































sui 
Ny} 
: 
NI 
N 
Ny 
NN 











Q07 

















| 












































0.010 
Ore 
0014 \H 
0016 
0078 
0020 
0.022 Wt 
0.024 
0026 
0028 


Values of p, Values of p, 
Fe= 800, fs = (6,000, = 12 Se = 00, 4s = 16,000, n= 12 























Q03 














of 


Q02 





Va/ues 








00) \—— Oe? 
PD WIALL 

















Values of p, 
Se=/000, F,= 16 000,17 =12 


p: = ratio of tension steel; p' = ratio of compression steel; a = ratio of 
depth of compression steel to depth of tension steel. 


Dracram 8.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel p:. 


(See p. 903.) 
Stresses in Concrete, fc = 800, 900, 1 000; 
Stress in steel, fs = 16000; 
and nm = 12. 



























































































eee 
& ~ aes) (aS Ket 
2 005 ee ee — aa 
aes Oe 
> =. : > aoa Ge Se oM, Vale ara ae oom 
+ (ran SS GRAS SSE RE Bee AF aa AS et kt — 
Q04 PF rf, US 003 ee oy po a as 
rT TF % ee ee ee aa aan 
fe} 003 Sy oe a ee he} eT a Re 
= —— t= 10? ft 
aoe Eee eS oats Geeta Een SS CS REA een PE 
Se er oe Ae ae pt Se ESS ae 
002 pt ey RES SE 
ee Ses ne comes 
ee ee ee ee Q07 eet ame ne 
00) Se a i a 
eo ed a Re eas 
iF iT ee a a a See as aS 
0 aS! eS ane tee 0 ELEN EE: ELE | ee FSG ee 
@DBDON WY SH & NY ODD SON wW SH DON KYO DWH 
So GDRVNRrRWRPTHANAWNAAD Serr tT evrnwnvWaun NY 
oe ogoouogeeao ogeogogaogaeese 8 Ss S S&S 
SseossssgsgsssSss SS8sssssegssqg 
Valves o Values oO 
i 1 
Tc = 650, fs = 18,000, n= 15 Te = 750, Js = 18 000,07 = 15 
OO ee ee eee 
(6 Q04 es SS a Ye a 
Ee ee eee ree Sa ae 
oases Sk RE ed Sh Me Be a 
Sara ee GS ee SD pf tf 
004 pene we Pees, 
ase ey Ses Pe Sea 003 eae a a 
Ea ees a ey," Be SSeS 
SS Gey GE es RI ae hs ie BS eS 
= ao Sa ee Es Va ar ccsenee enn] Se aaa| 
ean Tees ee ee HI LS ~~ Rei roe 
8 EE ES 
ue 0.03 oT AES a ee 
° eo ee ee Be } {|_| _}_4 Se 
BE nn ne Ee Ss ORAS Ol Cae Bee Q02 a a SS IP OGG OS AES 
Ease Sy OPN eS OLN LE ee me EES TIAL LAL LA eee 
ry a a A A = ey 
v pA et HOOD i I ee 
2 002-4] [GY eT nae EET Ss PML eee tt 
h eae ES ) Z 4 Se es ee [eee if 4 Pea as 
“ Se ce ee ae es 6 a ee 
=n eed j Sia ie ae ee A 
eS RW 0000 C07 BEE Ee A a ea ttt tt _ 
BE B90 75 8 A ES 00/7 a 
TNT a a ae Se jared Baer Bes 
Q0/ (ES /SU) 07h ES Ge es a) CE EG liSaeeses 
LK es SS SS as tee 3 
WLLL a ae Gas PS DE een pees [| 
ee Rw Se ee 
Ws eee eed Ee RS SSS ee | a ge | AY 
aes Ee Ree es Bere Se Pee 
as eee Grae es BS foe gah aM aes Oe ea 
ee a SS inane Bec ek od 
0 an SUS eee ON ee See 
SSssgsesss 8 os & = 8.8 8.5 -coeeees 
ssscososoosoo SSS os Ss S$ SS -o SS) sac 7S 
Values of -p, Values of pP, 


Te = 800, Js = 18 000, 0=15 Jc = 400, fs = 18 000,77 = 15 


p: = ratio of tension steel; p’ = ratio of compression steel; a = ratio of 
depth of compression steel to depth of tension steel. 


Diagram 9.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel pi. 


(See p. 903.) 
Stresses in Concrete, f, = 650, 750, 800, 900; 
Stress in Steel, fs = 18 000; 
and n = 15. 


0.07 


006 


Q05 


Qo 


of p’ 


Q03 


Va/ues 


O62 Ee 





i 























































= = 
Po aa a 4 
OFT 

















asa, 
e, 
=o 
a 
[| 


















































Set BE SIF a OF LF ASS. 








t-- fi Hh 4 Ci BF, 
soos e: oma a, Po 












t+— 5 vows A 








4 ve LA, 22 am 




































































= 











































































































007 


























v 
7 == 
STS NSW we So SOUS DO WD © SQ Ss 88 co FOr FON NS USOmS © 
Ss Se eS NO NN a Serr pe So OT BON CNN AG nD 
SSS 97S OSS Oo S S SS: S 
SSSSSECSCSECSSS SSSSSSEESSSE 
Values of P, Vatues of P, 
Sc = 800, fs = /8 000, n= I2 fc = 900,7; = 18,000, n=12 


Q04 


Q03 


CE Hs 


Q02 


Values 


001 - Be 


P1 


























= ratio of tension steel; 


























Va/ues 
F-= 1000, fr = 18,000, n= 12 


of P, 


Pp 





4 


= ratio of compression steel; a = ratio of 
depth of compression steel to depth of tension steel. 


DraGcraM 10.—Ratio of Compression Steel p’ for Given Ratio of Tension Steel p:. 
(See p. 903.) 


Stresses in Concrete, fc 


Stress in Steel, 
and 


fs 


800, 900, 1 000; 
18 000; 
12. 


907 


908 TABLES AND DIAGRAMS 


saa uea'iy 
TTT TAL VA 
HY] A La 





Al 
/ 
fi | 
( 
AA, 
aA 
r | 
V 








7407 a0 
002 AL 0.002 CI 
.007 010 Qdls a 02025 2030 0035 0040 .007 ae 2015 0. ma 2025 0.030 0035 0.040 
(SUG NOE a-0.04 











Ss 


U] 











































0.030 Hy Tern my .mney aunaay co 
Mh ra ro anny 408 
ty ACTA Po ELE ,Aanny ane 

1 Ht! ATO Coe Aun” nnne 
4o HL ra A es A 557400008 
0.040 Fr Ol 


im 

























ae 


A | f VY 
wi efi ‘alan l 
ew, a 
Ap dan sues 
0.030 el 7 71] WA ) 
v, (AV [IAT I 
wad | 


N 
= 














[| 
A 
as 








S 
x 
Ss 
S 
a 
A | 
NE 
Ch 











Katin oj Cornrpression Stee/ p 
















































































J ad 
4 fo 





0.050 5 TA A ney VY 
fifi Vy 
UAT A | iE 























HA, 
VY VIA VT 














PARA XL Yar: | | 
A AK 
Me pad 












AL 
Cae 
eae os 
tq ian 




































ooe 0.002 
007 0010 Q015 0020 aoes 0030 0035 a040 .007 0.010 001 


ie 
> 
S 
Q2 


2025 0030 0.035 0040 
a=0.1/2 


Ratio of Tension Stee/ p, 
a:Ratio of Depth of Steel in Compression to Depth of Steel in Tension 


Dracram 11.—Relation between Tensile and Compressive Steel in Beams 
with Steel in Top and Bottom. (See pp. 238 and 903.) 


TABLES AND DIAGRAMS 909 


0050 (TT 
Lt Alf aie 

















































































































































































































































































































































































0020 
Seat 55.057 dann bY eat 
LAT et LAT | a 
0002 a0 .Gn87.4074 0002 & Vit 
1007 0010 Q015 0020 0.025 Q030 0035 2040 007 0010 2015 0020 0025 0030 0035 0040 
a=0/4 a=0/6 ; 
0050 0.050 
CTT TAT) 2930 poo oo 
O60 A000 An) Gy Gey Oey Anne ar UariAF an ty Pian HY 
CO HA B) (70 AB 007 S08 G08 SER nee’ A 
PA 7 BA) a7 A645 4NbY Suan a “au 
odo Hef ATT oo DO 4 
rl S00 AS i sn ann y, 
ry PY PVA ot bP I] au 
aie any aERy. 7; 7} ) 
HANAFI) Acerca cous V28 
0.030 1H BY, etd 4 0030 4 
10 aw Ny SERED CAERUD SERB pa 
PTY PAT TT] rT 
AULT ATTA TAT 
181940407 G00 .00"9/4008 pin Paicae 
0020 FA fr! go20H SEGEe 
wl aReae 
: AHH 
f a Rath s 
0070 WHE 4 0070 ReaaR 
LA 












































Paul 
2002 Ae HH Z BECEEH EHH HEH 
007 0.010 0015 A020 0025 0030 0035 a040 .007 2010 0.015 0020 aes 0030 2035 0040 
a=018 Ga. 





















0.050 
















































































































































































































nEEEAS TOMO 
nan/08 CC i 
smnuen/ ann aang , 
55/000007,0008 7 
GUSnny dene 0040 np48 
ARGO) 400888 1040 ay 47a 
4 C 
f) 
a wos Ht 
i Ho ay 4a6 
ue aan 
1m 0407000 408 a y, 
0.020 FI 0.020 anP74 
An An Ya 767 07 G00'4020 488 
HAT ATTA aT 
HEA Hy 
0.010 Fee 0.070 TET 
PLA ee I PLL EY TT 
edy Gay aan TT ALITY SUScROnG8 
AvGV Ann 4058 Hc 2 ALEC 





2002 0.00. 
-007 0010 0015 0020 0025 0030 0035 0040 907 Q010 Q0IS 0020 0025 0030 2035 0040 
a= 022 a=0.24 
Ratio of Tension Steel p, 
a=FRatio of Depth of Steel in Compression to Depth of Stee/ in Tension 





Diagram 12.—Relation between Tensile and Compressive Steel in Beams 
with Steel in Top and Bottom. (See pp. 238 and 903.) 


TABLES AND DIAGRAMS 


910 


on 
hy 
oe 
gia 6 





a 0.20 0.22, 0.24 





























G fo sanjog 


Js 
nfo 


Values of 
Ratio of Depth of Steel in Compression to Depth of Steel in Tension. 


a= 


—Values of j for Beams with Steel in Top and Bottom. 


(See p. 240.) 


DracRamM 13 


TABLES AND DIAGRAMS 


911 


USE FOR FLAT SLAB DESIGN 





003 

















aa 


Constants C4 Cs Ce 


0.025 7 




















S 
S 
No 





























Constants Cy Co C3 








a 
Sold lines Jor {,=16000 lb per Sgn. 
Dashlines for f,.=18000 Ib. per Sa. in. 





0.07 
S Sey 
Values of :: 


of Flat Slab at Column. 


0024 





























cor LLL RN 
SERGE TERRE RRRSSe! 

Me sla Le Hebd ET AEP 

eee bh a] f 


Solid Ines for f =16000 /b per Sq. in 
Dash lines for 7,=18000/t per Sq.in. 











0.04 


O02 





Ml 
Values of r 
Diagram 14.—Constants C1, C2, C3 and C4, Cs, Cs in Formulas for Thickness 


(See pp. 338 and 339.) 








7 





























0.078 nei 


0.076 


Constants C 


0.0714 





0.072 





I = 
CEE CSS Sct 
Mica back helt 











>) (LESSER eee 
fon) - rol 
= N N ON 

Values of i 


Diacram 15.—Constants C;, C3; and Cy in Formulas for Thickness of Slab in 
Middle of Panel. (See pp. 340 and 341.) 


912 TABLES AND DIAGRAMS 


Table 17.—Flat Slab Constants Cio and C,; for Computing Compression 


in Concrete. 
Values of Cio 
Cc 
0.28(1 _ 1.28) 





























Cy = 7 in Formulas (29), (31), and (82), p. 344. 
a 

a 

c 

i 0.1 0.195 | 0,15:120 Avo) aOeee 0.225 | 0.25 | 0.275 

| 
C10 0.647 | 0.625 | 0.603 0.581 | 0.559 | 0.537 | 0.515 0.493 
| 

















Values of Cu 


b 1 bY fa \4 d : 
Cu = i +. ( — *) (“) (2.55 ~ 1.5) ji Formula (31), p. 344. 


ee ES fl .tttttC<C~stsSS 


b 
Values of — 
Li 


ee 


Sle 


0.2 | 0.25 | 0.275 | 0.3 | 0.825 | 0.38 0.375 | 0.4 

0.1773] 0.2311] 0.2580) 0.2849} 0.3118) 0.3387 0.3655| 0.3924 
0.1813 0.2344| 0.2610] 0.2875] 0.3141; 0.3406 0.3672| 0.3938 
0.1887] 0.2406] 0.2665] 0.2925] 0.3183) 0.3444 0.3703} 0.3962 




















0.2158] 0.2632] 0.2868} 0.3105 0.3342} 0.3579] 0.3816) 0.4053 
0. 2367| 0.2806] 0.3026, 0.3245 0.3464] 0.3684] 0.3903} 0.4122 
0. 2632] 0.3027] 0.3224] 0.3421) 0.3619 0.3816} 0.4013) 0.4211 
0.2960! 0.3300| 0.3470] 0.3640 0.3810) 0.3980] 0.4150} 0.4320 
0.3353] 0.3628| 0.3765; 0.3902 0.4040) 0.4177| 0.4314} 0.4451 
0.3822] 0.4018] 0.4116] 0.4215) 0.4313 0.4411} 0.4509] 0.4607 
0.4367 eae 0.4525] 0.4578] 0.4631) 0.4684, 0.4736 0.4789 


OCOOMONNMOSAAE 
nAonononoao uo 





| 
| 
| 
| 
| 
| 
| 
| 
| 
| 


ee) eae) (e) Fae) (eerie) a ea i=) aio aKa ia ani 


S 
(=) 


ooo © 
bo He & OO 


(ooo) 
oo 


“I 00 00 © © 
maon rf & 


Qnty an 
CONK OD 


OD 
H OD 


a 
So wl 


TABLES AND DIAGRAMS 


Table 18.—Flat Slab Constant C,, for Computing Shearing Stresses 
at Column Head. 


OF 


OS OF Oro OSS aa oro Crore" oo ore 


16 


.185 
. 182 
iS 
174 
.170 
aye 
.163 
. 159 
.155 


151 
148 


.144 
. 140 
. 136 
. 132 
ae 
.125 


121 


1I7 
.114 
.110 


02 


eile le eee) eee ea eae eS ei) ee ee eo m=) 


18 


. 164 
.161 
.157 
.154 
lied 
. 147 
. 144 


140 


. 137 
.134 
. 130 
Sar 
.124 
- 120 


117 


.113 
2110 
s107 
.103 
. 100 
.097 


0. 


2) See eee) Se) ea es Nea es) (es) (a ta) (a iS ye eal is) 


20 


.147 
. 144 
.141 


138 


.135 


131 


.128 
.125 
.122 
9 LARS, 
.116 


113 
110 


. 107 
. 104 
SCL 


098 
095 


.092 
.089 
. 086 


0. 








SO OVO FOr orerSrS OVOrSr Oro OrocOreoreo,Ooie 7S 


22 


.132 
. 130 
127 
.124 


121 


lhe 
.116 
peliies 
.110 


108 
105 


.102 
.099 
.097 
.094 
.091 
.088 
.086 
.083 
.080 
O77 





l 


in Formula (39), p. 349; v = Crow. 


Values of Cie 
ee a 


0. 


.120 


.115 
.113 
.110 
. 108 
.105 


. 100 
.098 
.095 
.093 
.090 
.088 
.085 
.083 
. 080 
.078)0. 
.075)0. 
.073)0. 
.070)0. 


CRO LO LOS s Orr So S20 Oreo ororOiS Oreo) 


Values of - 


24 |0 


118 


103 

















0.30 | 0.32 | 0.34 





SS Sees S&S Svea eS = 





Siete) eee) eee eae) (Stee a= ke moaonr=) 





| 
| 


SSeS TS SOLOS Sioreoreioo orovoso.So SoS 


.081 
.079 
s077 
.076 
074 
.072 
.070 
. 069 
. 067 
.065 
.063 
.061 
.060 
,058 
. 056 
.054 


.052} 


.051 
.049 
.047 
.045 

















0.36. | 0.38 
0.076|0.071 
0.074|0.069 
0.072|0.067 
0.071|0.066 
0.069 |0.064 
0.067|0.063 
0.065/0.061 
0.064/0.059 
0.062/0.058 
0.060|0.056 
0.059 |0.055 
0.057 |0.053 
0.055/0.052 
0.054/0.050 
0.052)0.048 
0.050/0.047 
0.049)0.045 
0.047|0.044 
0.045/0.042 
0.044/0.040 
0.042/0.039 





913 


914 TABLES AND DIAGRAMS 


Table 19.—Flat Slab Constant C1; for Computing Shearing Stresses 
at Edge of Drop Panel. 


(i) 
oe 
l 


l 
Cis = in Formula (42), p. 349; v = Cisw 





Values of C13 





C3 
Values of a 


























0 .078|0.072|0.067 
0.077|0.071|0.065 
0 .075|0.069|0.064 
0 .073|0.067|0.062 
0.072|0.066|0.061 
0.070|0.064|0.059 
0.0680. 063|0.058 
0.066|0.061|0.056 
0 .065|0.059|0.055 
0 .063|0.058|0.053 
0.061/0.056|0.052 
0.060|0.055|0.050 
0.058|0.053/0.049 
0.056|0.052|0.047 
0 .055|0.050|0.046 
0.053|0.048|0.044 
0.051|0.047|0.043 
0.049|0.045]0.041 
0.048|0.044|0.040 
0 .046|0 .042|0.038|0 
0.044|0.040|0.037|0 














TABLES AND DIAGRAMS 


Table 20.—Properties of Column Sections. 


Areas, Volumes and Moments of Inertia 

























































































915 


a 


Volume! Moment 


of 
Inertia 





In.* 





1 728 
2 380 
3 201 
4219 


5 461 
6 960 
8 748 
10 860 
13 333 


16 207 
19 521 
23 320 
27 648 
32 552 


38 081 
44 287 
51 221 
08 940 
67 500 


76 960 
87 381 
96 827 
111 361 
125 052 


139 968 
156 180 
173 761 
192 787 
213 333 


Round Columns. Octagonal Columns. Square Columns. 
Dia- | 
meter Volume|Moment | Volume} Moment 
d Area | per of Area, per of Area per 
Ft. | Inertia Ft. | Inertia Ft. 

In. | Sq. In. |Cu.Ft.| In.* | Sq. In. |Cu.Ft.| In.4 |Sq. In.| Cu. Ft. 
12 113.1] 0.78 1018; 119.2) 0.83 Iets6 ie 144 1.00 
13 132.71 0.92 1402| 140.0) 0.97 1565! 169 i Nog Be 
14 153.9} 1.07 11886) —162..4/-1.12 2105} 196 1.36 | 
15 | 176.7) 1.22| 2485) 186.4, 1.29] 2775] 225] 1.56 
16 2011) 1.40 wali 2i2. 17.1 Az 3591) 256 1.78 
LT 227.0} 1.57 | 4100) 239.4! 1.66 4577| 289 2.01 
18 254.5) 1.77 5153} 268.4! 1.86 5 753| 324 2425 
19 283.5) 1.97 6397; 299.1) 2.08 7142) 361 Ler 
20 314.2) 2.18 7 854' 331.4! 2.30 8 768! 400 Zate 
21 346.4) 2.40 9.547). 365.3) 2.53 | 10658} 441 3.06 
22 380.1} 2.64 | 11499) 401.0: 2.78 | 12837] 484 3.36 
23 | 415.5] 2.88 | 13737 438.2 3.04 | 15335] 529] 3.67 
24 452.4; 3.14 | 16286) 477.2) 3.31 18181| 576] 4.00 
25 490.9 3.41 | 19175; 517.8] 3.59 | 21406! 625 |] 4.34 
26 530.9} 3.69 | 22432; 560.0) 3.89 | 25042) 676] 4.69 
27 572.6) 3.97 | 26087! 603.9] 4.19 | 29123] 729 5.06 
28 615.8) 4.27 | 30172, 649.5) 4.51 | 33683] 784 5.44 
29 | 660.5] 4.59 | 347191 696.7 4.84] 38759} 841] 5.84 
30 706.9) 4.91 | 39761; 745.6) 5.17 | 44388! 900! 6.25 
31 754.8} 5.24 | 45333] 796.1! 5.52 | 50609} 961 6.67 
Ba 804.2) 5.58 | 51472; 848.3] 5.89 | 57 462} 1024 | 7.12 
33 855.3) 5.94 | 58214; 902.2) 6.26 | 64 988! 1 089 7.56 
34 907.9| 6.30 | 65597} 957.7) 6.64 | 73231/1156 | 8.03 
35 962.1) 6.68 | 73662) 1014.8) 7.04 | 82234/1225 | 8.50 
36 |1017.9) 7.06 | 82448) 1073.6] 7.45 | 92043 1 296 9.00 
37 |1075.2| 7.47 | 91998) 1134.1] 7.87. | 102 704! 1369 9.50 
38 |1184.1| 7.87 | 102 354! 1196.3] 8.30 114 265/144 10.02 
39 | 1194.6) 8.29 | 113 561} 1 260.0) 8.74 | 126777| 1521 10.57 
40 1 256.6) 8.72 | 125 664; 1 325.5} 9.20 | 140 288! 1 600 | pa ee 



































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OIF I 8ZE I 6&2 T OST T 
09g I CLZ I O6I T COL I 
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91Z I 961 I FLY 1€0 T 
QFz I OLI T Z60 T FIO T 
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Z6L I SILI eF0 TL 696 
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= ZGO T 986 1Z6 6G8 
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la 896 806 L¥8 18h 
0F6 T88 E78 FOL 
a Z16 ecg S6L aa 
a $88 628 PLL SIL 
a (a) (01) (6) (8) (2) 
BS 00g = F | oe =F | 002 =F | 099 = 
~Q ‘ 
<i roi 
= uj ‘bg red ‘qT ul suumnjod) uo 


FOL T ZIOT 026 8Z8 982 0090°0 
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926 8o8 OSL OL $29 00F0 0 
C16 6E8 £92 989 O19 G20 °0 
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(9) (¢) (¥) | (g) (Z) (1) 
009 = 2 | 06¢ = F | 008 = 7 | .0SP_= | OOF =F 




















d [3949 jo 0198 
‘f ‘ssa11Q JUL], 9BBIOAYV aIqVMO]LV 





cl 


U 


APOYSTLG 
jo IMpoW 
jo ORY 














Coop -d 20g)  ‘[d(t — ¥) + IPS = f uo posed 


-J90}8 [BUIPNYBUOT Jo gaSeyusor1ed yuoIOyIp puv 9301009 UI sossodjs FUN quoleyip 10,7 


“{ ‘sasso}G BUIyIOM a3eIoAy—'SuUIN[OD 93}919N0) poolojulay— TZ PGB L 


916 


1A 


TABLES AND DIAGRAMS 













































































F69 I OF I €9F I 98¢ I 60€ I BES I ccT I 820 I 000 I £36 00900 
Cro I C6F I OZF I CVE I 123 1 961 T IZ1 I Lv0 I oL6 168 0¢¢0'0 
c6¢ T OSF T £ge0, COE I BES I O91 T 480 T C10 IL 3F6 0L8 00¢0°0 
OLS T SZF I 9¢€ T C8z T €1z I SPI I T20 I 666 826 LS8 ¢L40°0 
9F¢ I cor I cee I b9Z I v6r T vor I rS0 I b86 S16 €F8 0S+0'0 
Gze I ESE I eI I vvE I GLI I 90T I LEO I 896 668 0€s SZr0'0 
96F T 09€ T 262 T b2e I 9ST T 880 I 020 T 396 +88 918 00700 
ILh I SEE I TL I b02 I LE TR], O20 T €00 T 966 698 £08 ¢260'0 
LEY I cIg I 6FS I S8T I SILT 290 I 986 126 ccs 682 0¢S€0'0 
GZF I $66 I 832 T SOT I 660 I PE I 696 £06 OF8 QLL S&E0 ‘0 
L6& T O22 T 902 I EFL I 620 T 910 I £96 _ 688 928 BOL 00€0'0 
SLE I 8FZ I SST I E21 I 090 I 866 966 £28 IT8 6FL G2Z0'0 
SPE I C3S I vor T SOT I I¥0 I 086 616 8¢8 962 Gel 020 '0 
EZE I 602 I orl I 680 I 330 I 296 306 ag) 68d BSL GZZ0'0 
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E26 I SST T OOT T 3rO I +86 926 898 O18 ood G69 GL10°0 
6FS I CsI I 820 I 620 I G96 806 1¢g C6L 8E2 189 OST0'0 
b2z I eri 290 I T00 I 9F6 068 ves 6LL €SL 899 SZ10'0 
661 T 060 I ¢€0 T 186 926 318 818 €92 602 bg9 0010°0 
bLT I S901 -  FIOT 196 406 eS i 108 LYL +69 1¥9 ¢200°0 Or 
OOrT = “f/ 000T =| 096 =F | 006 = % | oss = ¥ | 008 =F | OG2 = F | 002 = ¥ | OC9 =F | O09 = F 
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O€s I 696 I L6T 1 O€I T 490 I 866 T€6 ¢98 862 bel 00€0 ‘0 
€0€ I LEZ I GLI I LOI I 6t0 I LL6 316 L¥8 BSL OTL ¢220'0 
C1Z I 11 T LPL T 80 I 020 T 196 €68 628 GOL TOL 0¢Z0'0 
8FZ I SST I €ZT I 090 T 866 966 E18 TiS) dss GEL 989 6220 0 
022 I 6ST I 860 T L€0 I 926 C16 bos €6L BEL 129 00z0°0 
S61 I Sete €20 T PIO T +96 +68 ces GLL 912 9¢9 GZ10°0 
SOI I LOT I 8r0 I 066 366 v8 918 LoL 669 +9 0ST0'0 
Ss I 180 T $60 T 196 O16 €¢8 96L 6&2 €89 929 GZ10°0 
OIL T ScO T 666 £F6 888 S€8. LLL CSL 999 119 00TO'0 
een ee peony rL6 026 998 o18 Sol FOL | o¢9 G6¢ ¢200°0 oI 
OCOT = “/| OS6 = /% | 006 = ¥%-| OS8 =F | 008 =F.) O94. =F | 002 =F | 060 = F | 000 = Ps | Ogg = AF 


918 TABLES AND DIAGRAMS 


Table 22.—Spiral Columns—New York Code. 


Values of fi = ap 1 (n = Lp 2» | 


in formula, P = Ate +(n—1)p+ >» | = Af,. (See p. 424.) 


fe = 600, fs = 20000, n = 12. 


Values of fi 


eS eee 








Ratio of Ratio of Vertical Steel p. 
Spiral 3 
| 
HY 0.01 0.015 | 0.020 | 0.025 | 0.030 | 0.035 | 0.040 




















fe = 500 J, = 20 000 n = 12 TN. al ee ee 





0.01 970 1 005 1 040 1075 1110 1 145 1 180 
0.0125 1 070 1105 1 140 1175 1 210 1 245 1 280 
0.15 1170 1 205 1 240 1 275 1 310 1 345 1 380 
0.175 1 270 1 305 1 340 1375 1 410 1 445 1 480 
0.20 1 370 1 405 1 440 1 475 1510 1 545 1 580 





fc = 600 fs = 20 000 n = 12 mix = bilgo3 


Sepa eu ERI cck ec 


0.01 1 066 1 099 1182 1 165 1 198 1 231 1 264 
0.0125 1 166 1199 1 232 1 265 1 298 1331 1 364 
0.015 1 266 1 299 1332 1 365 1398 1431 1 464 
0.0175 1 366 1 399 1 432 1 465 1 498 1531 1 564 
0.020 1 466 1 499 1 532 1 565 1 598 1631 1 664 








Note: For spiral design recommended by the authors average stress f may 
be formed from T able 21, p. 916, for appropriate values of f,, and tables on 
pp. 926 to 928 may be used for design. Pp. 929 to 934 give data applicable to 
all spiral columns. 







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to ea es age ee ee , STs SS \ OTS SS Bae: eee ee ER ees Cae ee mer ty eres ee ey ee Ee 


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Column Loae 
Based on Formula 






.—Total Loads on Round 






















































‘| 
es A ee eee ee ies ee Ol ae eee Bey oe ee eee 
oe a Sa, Tee 's, SE Sats. SE a. Wi ee ee, ee TS, Gres, “i ST Se RRR TS SERS ns prs Sieg gig eS | ee Ge CE 
SS \o ee Ey ‘i a A ee SE HW, AE hc | Se Bs rs ee Sees 1 ee Ses eS Se ee 
a a it eae ee Te, Se ik a eee Se, WE. ES EA) S| ee ee ae" ee | ed Ee SE ee ee 
ee et So A ee ee a ee ee ee i a es ia ee el ee ee ee ee ee 00 9 ive) 
a * 6 Ee ee ES ae. OS, WS RE 8, E.R ee" | a ee ee ee 4 
iso ae ST S te Le eee ee ee ee eee a Oe ee ee ee ee ee ee 
te Lee ee ee ee ee ee SS Ss 
ca a a = aa Yee ee ee ee el ae a eee ee ee on a ee ae ee ee a 
REIS - Sale ee rae ae oe ~ GES Ea a a a TS RN I, te, “Ge SG, HE. Gat IR, WSS SE es SS ee  -  e Ieee eae aes pe 
ot de I 5 a SS as Oe ek ES a, a Ee, Ea G ea WS | RE Pe ee eS Ee GS 6 ee eS 
Pe ee (ia, Sa mee, ea Re Se ea ene. Gea Sf Rae ee RS eS eet a a aS RN eet S 
aa ‘a, ‘ee BOR idle. -O) Aneesh RS ei SS, SS». SS oh SP ee De Be eee 5 Se aa om ae < 
Te eo Se SEE BOE Le St SORE Bee Sse 1 72 RO GS a, SE, Gee on BS oo a ee ee Ee 4 
ae (i Se tas bh Gh SSeS Se OE Se Wg), OE Ee ee) ee Re Ge 00 G Q 
— | A ee Se eS a ee aoe 
Si “nod ee eens ie a. Es SS <a eS. re 
% Se EAS ThE SS ee A ee ea, Se I a ie ar a ae a a ae ee eT eee oe Sk ce ete et 
a a ha ee aE ee SO a ES Sa 
N st Giles a > 
= iS ts ER ORT Pant ee for MR oe ee 
ies 1 en ee DO ES SR eS RE ee, 
as Gh GHG aire Pee 1 er 5 a Bes Tee eG ee 
——f SS Be he SSR 2), AE es - e eee S 
in ceo ee Sn a oe 0OOr 
CS ae ee a ee 
ame. ~ (ie, A. US, i, GR ER ee ee ee 
: 1 GREE, \L_3S. IEE, le ee ee ee ES 
s $2. SE ROE. Se, CS ae 2 ee ee Ee Re 
1 SS (TSS BD, Th OD als LS RE 
a (Sk I ke, Gag ee Pe ee I ee Be 
.o SS SL ON ee ee ee ee 
CAS See SS 
a Se ee ee ee ee Se 
ce, RE RT ii. ee, a2 a ES OlORS 
| eee eee ee ee SI ei tT 
be SG ML. SR, SE ENS, he GE ee ee eee 
=~ fees Se, SS SS Rly, a Gi, Sl SS ee eee 
SEE SIMs GS |. Ee, I, Se. el SG a, ET ee ee 
1 EES ae RR CR, NU, SR sy ON OE Gere 
= eS ES TSE ae ee 
eae ee eR a i Sh ee ee 
SS a ES En, ee ee 
a eS Se Nee 
See eee eee en ee Se | 00 é 
as SS ee es Ra a A i SEE FE RE Be RS ES, ES, Ee 
= a eae ee See ee a Ce ys RN Se , G, , , SR eet Eee 
es oe es ce a Se ie, Sa eS ee 
See Lip aeaeee ee er ee ee Se ee ae ee ae a | 
rae (ae es SS Sa ae a ee ES SE ee eS, Dy RE 
es aa Soars | (wenn ieee) Ge Da ee 2 Re GE MRS Be Ee ES ES 
EME Sate ESS TS w= CT | ee i Si Sh. Ss 
Gees aaa GN SS SET] eae) (GEE 8 (eae TR te a SSS SSS SSS 
SS ae es ets Re FS a he 
as ess Sa, Se a ee Sead ee ee —— =| SS SS Se 
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Fffective Diameter of Round Column. 








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S 
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Column Load 


Diagram 17.—Total Loads on Squari 
Based on Formula 














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PRL nN 
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MAL rail ie vail 


HI CTE TE Tt 

eat ieee ec 

ll ll ie ail sect AANANNNUAN UL 
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HLA ATATALTETHELA A ETLALAT HELL AY THT 


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Wildes ER Se, Se LY +4} 


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HHA AEE PUTTU TIPE 
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1/000 Pounds 
imns for Different Average Stresses f. 
Af. (See p. 406.) 


(To face page 919.) 


Effective Side of Square Column, in Inches 


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TABLES AND DIAGRAMS 919 


Table 23.—Spiral Columns—Chicago Code. 


Values of fo =fe(1 + (n — 1)p + 24np:) 
in formula, P = Af,{1 + (n — 1)p + 24npi] = Afs. (See p. 425.) 


Values of fe 


Ratio of Vertical Steel p 






































Ratio of 
Spiral | | 

p1 0.01 | 0.015 | 0.020 | 0.025 | 0.030 | 0.035 | 0.040 | 0.050 | 0.60 

Jee=-12d Ib. nm = 10 mic bees 2 
l 
0.005 881 | 913 946| 979] 1011 | 1 044 | 1076 | 1142 | 1207 
0.0075 926 959 991 | 1024; 1057; 1089) 1122 | 1187 | 1252 
0.010 971 |1003 | 1037 | 1069] 1102] 11384] 1167 | 1232 | 1298 
OLZS” lero. 1049 | 1082) 1115} 1147| 1180] 1212 |,1278 | 13438 
ROLO ee tle oa 1095 | 1127} 1160} 1192] 1225) 1258 | 1323 | 1388 
OS ONES Ts ae aed I eae 1173 | 1205) 1238] 1270) 1303 | 1368 | 1434 
aa Us aie bla amo 1 218| 1250| 1283) 1316| 1348 | 1414 | 1479 
fe = 600 n = 12 mix 1 :13:3 

0.005 756 | 789 822} 855) 888| 921 954 | 1020 | 1086 
0.0075 801 834 867 900 933 966 999 | 1065 | 11381 
0.010 846 | 879 912} 945; 978] 1011] 1044 | 1110 | 1176 
0 ae ce 924 957 990 | 1028; 1056) 1089 | 1155 | 1221 
Je eee 969 | 1002} 1035}; 1068; 1101| 11384 | 1200 | 1266 
Me dey. SP em brs 1047; 1080} 1113] 1146] 1179 | 1245 | 1311 


J 228 Eid Rea eee ea eee 1092) 1125) 1158); 1191} 1224 | 1290 | 1356 


Note: For spiral design recommended by the authors average stress f may 

be formed from Table 21, p. 916, for appropriate values of f- and Tables on pp. 
926 to 928 may be used for design. Pp. 929 to 934 give data applicable to all 
spiral columns. 





TABLES AND DIAGRAMS 


920 





























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TABLES AND DIAGRAMS 















































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0F0'0 =d cs0'0 =a 0s0'0 = a cc0'0 = 4 020'0 = d@ cI0'0 = ad 010'0 = d 























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TABLES AND DIAGRAMS 


922 































































































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“ul ‘bs |"q] OOOT| “Ut ‘bs |G, OOO T| “UE “bs | qT OOOT ‘ut “bs |"q] OOOT| “Ut ‘bs |'q, OOO T| “GES | GI OOOT| “UE ‘bs |"q] OOOT “ul ‘ul “ul 
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eee as * 
ans GIPIMm 4pt 
oro'0 = @ cg0'0 = & 0g0°0 = @ ¢Z0'0 = @ 0z0°'0 = @ c10'0 = a 0100 =4@ seta as 















































949I1NU0D Jo BTV SATOH 0} [9999 JO Bory JO ONVY 











(90F *@ 299) 


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Aju Woursss10yuIeY eoseA YM suUINjOD arenbsS 9% 9IGeL 





=_—T a 


923 


TABLES AND DIAGRAMS 












































































































































C°SE | -9249 | 4°88} ShO. | GSS FUSLIOLGS TL bo ese [eS er’ Seen ba eli) veg em 6 F6F cg 8e 68 
€°OS-| 289 +) 8°18 | 809. C12 1 080. | £ oo 1 leu Rl fee =) Gein ope be COF FE le SE 
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2 06°} O89 <> & 92 | 90S (1 O gec CSP |G SE le GUr IPE Gkal oy. 1.0 tee reo liy ieee L8€ yes re ce 
C82") 06h of LIPS | PLP) 21S | SG eT | Oey oe Pe 20 OO OL | cee for £98 og ge rE 
02 \ POY 4 Eee Ser Ber | ecr  er ort eipy 1 Goer. tse ol Gre ose | 9°9 6ge 63 Ze ee 
QiG! Ser | Oe b> Cle. Goat SOS a Ren PLC a Oar he Oe eG cee | 29 91g 8z 1€ Ee 
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Salo) Lee Ce EL See Pe eee eG 125-902 Ohara. LOZ~ + 8°S S61 ZS GS 9 
Biel eye Bi inl eee Poy OLS bee ee OIZ | 69 002 | 2°¢ 68I | ¢'e SLI 13 ¥S 8 cZ 3 
OCT Notes UAL. Me eee 10Z0 4 Se TEL 8-9 | ee a (i a a 19 0z eo 4 3 
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T's gg LZ e¢ oS o¢ 0'% SF 9° CF ZT ef 8°0 OF Ot el ¥L 
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0% ce ST bE cI oe Sab og OT 63 8°0 13 G0 9% 8 II ZI 
cl ge, Sot 9 BE c% OT &% 8°0 oS 9°0 1Z C0 ie L OI IT 
(UOTYwpusUIWIODeY ,SIOYINY) GL =U OSGF = % 93010U0D F: 7: T 
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sy d Bd d = | d Sy d °V d a 2d d “¥ d 
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fe | BOOT, po Reid pers] BOOT |e be Bete eae Beem ee peat oP| Won | Seamer | ie | nats 
BIIV | BsolV i Bolly ey Boly ae hE ’ BI1IV ; e: VATPOO HY I9zJIUL BIC, JoJIULBICT 
~ 0700 = 4 cso'0 =4@ | of0'0=4 c70' 0 = 4 ozo0=4@ | sloo=d o10'0 = 4d 
: 9JeIOUOD) JO VolY OATIOIYY OF [9040 JO BoIY JO ONVY | 
(‘90P “d aag) [d(T — 4) + Thfy = d Uo poseg 




















peoy UoATr) 10} pormboy ]90}g puv sazig snoleA JO suMINIOD 10J sBuIpeorT oyeg 
AJUC JUSUsdIOJUIDY [VIIA YIM suUINjOD punoy—'1z e1qeL 


TABLES AND DIAGRAMS 


924 































































































¢ ee) O62 | 288 | O92 | 683, 66L | 1 FE | G69 | @ OT 699 | VFL} OF9 | 96 609 GE 8& 
eo¢ | OF, 1 SIE | £1Z.2\-6 £e-1, 880. | 4.68,| 2099-1 i8t 129. | 9st |. £09. | LG PLE rE Le 
zee | ZOL | 66S | G29 42.98 | GPO? [Pr 1Sh| © Gad yt eT ceg | 8'sI | 89¢ | 98 IP ee 9€ 
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zoe | ozo | r'9e | 969 | 9°22 | S24 | 6 BL] SPE | IST eggo } etit | 10g. |) Sk SLP Ig rE 
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$92 (8 geo. | Pree} s tee, | 8 OR Oe. Te Bhs OShe | ee 6St | 66 6eh | 9°9 SIP 63 oe 
Opel 20g, | Sie) oer] Saske 20r, | FSi le ere | ee) Scr | 3'6 60h | 29 068 8z 1g 
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z1z | cep | 9-S1-| 61 | 6'SI| ZOP<)| S81 | S88] 9 08 | -.698 08 ase aed as CEE 9% 63 
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9°OL | <FR | SiFT-| $2@ | $:ZIu| S18] POL) 208.) © 8 68 | 3.9 91. | oF £92 &% 92... 2 
2°OT | SIS 1s Stl? 008, | 7 le). Seenunage 916 | OL 796. «|/L°S ZOG 828 OFS | 2 ae 
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Sy d oy d A d Sy d PVs d Sy d oY d 

994 9046 994 99049 9919 9949 39099 uuInjoy jo uuin[od 
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Bolly eS Bolly FPS Bolly eS Bolly eS BolY eS Boly BS BolY eS dATVOOHY I9jVoOUIBIC, 

0r0' 0 = 4 ¢e0'0 =a 0¢0'0 = @ cz0'0 = 4d 020°0 = c10 0 =4 0100 = 4 

















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(907 “@ 999) 


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925 


TABLES AND DIAGRAMS 
































































































































9}9INUOD JO BaIV IATPIWY OF [9049 JO Bory Jo ONBY 





(‘90F “@ 2a9)) 


‘[d(t — u) + 1Pfv = qd vO paseg 





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0r0'0 = 4 ¢e0'0 = 4 0s0'0 = @ Gz0'0 = 4 0z0'0 = 4d c10'0 = 4 o10'0 = 4 





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PpBO'T USAT OJ pormboy [904g puv sozIg sNolIBVA JO SUUINTO*D JOJ SSUIPBOT OFBG 


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TABLES AND DIAGRAMS 


926 






































L' LE \O8ZI|L' SF (SPIT [€ €F |S60I |G 8E JOSOT |Z EF OOO T 168% [966 |IT' FZ |606 |Z 61 \€98 |F FI |FI8 9°6 \992 
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(UOL}VPUIUIMIOIIY sioyyny) ST =% O0L = af ajelvUOD Fi ZS: T 

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a d ey d oY; d SV d Ag d SV d i d oF d vy d Sy d 
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0900 = 4 | 0900 = 4 | 400 = 4 | 0F0'0 = 4 | $800 = @ | 0800 = d|¢z0'0 = 2 | 020'0 = 2 | S100 = @ | 010'0 = @ 























































































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Jesids % J pue JUewedIoyUIOY [VITO A YIM SUUINO) jerds punoy ‘O§ 2198L 


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927 


TABLES AND DIAGRAMS 





















































































































































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TABLES AND DIAGRAMS 929 


Table 33.—Column Spiral—Forming One Per Cent of Core. 





Diameter of Wire 












































Diameter 4a aS TD 1 In he #1 
of Core, e : ane bi le 
In) | | 
Pitch,| Per |Pitch,| Per |Pitch,] Per |Pitch,| Per |Pitch,]| Per |Pitch,| Per 
Ine Cent jing | Cent | in: |‘Cent | In. |-Cent. |. In. | Cent | In. | Cent 
8 27 1.09 Bee 1.02 
9 2 1.09 34 1 1.05 
10 Po-oe 3) |1.02 | | 
11 12) 1.02 22 | 1.01 
12 toe L209 Diet lO 2, 
13 1$ [1.01 | 2b | 1.05 
14 ibe ede le 23 | 0.98 3 Oo 
15 i od ee? | 1.02 | 3. | 0.98 
16 14 | 0.98 13 | 1.10 25 | 1.00 
AE ZE RRR ES i aN Tei O38 DL Oe 
{SEE ee a 12, | 0.98 23 | 0.98 
Opera ell Pats ly. Sct ey 12.11.08 22 | 1.03 3 1.06 
| LR eR ene lem ee Os 2; | 0.98 3 1.00 
ee Ta erat melee uni ah 12 | 0.98 2 1.05 D7 anton eo! 
A aaa een ea ra Dele del 1 2 1.00 22 | 0.99 
ORT Hires SRA ae ae 1Z | 1.07 2 0.96 27 1 £205 
CALE Fo rath UR ih. ee 1 | 1.02 105 227 11.00 3 1.09 
AS ae NEE eh ae 1% | 0.98 ee tO Dare Mt Mel Ove 3.) i) 1.05 
225 el NRO ST a a Es a 12 | 0.97 27 1203 3 1.01 
ee OS on I atta eae ee A 13 | 1.09 21 | 0.99 3 0.97 
Pie Ml DR RR re er a Te 1.05 27) 70.95 OS alta 2 
OME Tanne Ee, NON Ris || oe, ches ws A ae, 2 1.04 22 | 0.99 
DOP re clk: iene alee, es eae ae 14 | 0.98 2 1.00 24 | 1.05 
foe SOG eG ae a 3 OF07s wagowele £01) |v 
i NSS ee i a 12°) 1:08 | Wey )6.98.|\ 3] 1104 
Po RN ee ee Mi Ceee Pee lave che ssf’: folds I's Shes 12 | 1.04 2% | 1.06 3 1.00 
a Se mene ne cme set ten | ong Piast a6, Sites | ee ceo ae 8 12 aie leow 24 | 1.03 3 0.98 
ort eens lee ee te fee a. S('Gat ats, Sail atode 3, iv secs. ace 13 | 0.98 21 | 1.00 22 | 1.03 
BAS! 8 yp oot lhe ie teas tena ian (2 Ae eed ote ae a 1 el leet 21 | 0.97 23 | 1.00 
5 PE etna orcatyc Were ae tre [one ee ies] tas 399 1%} 1.08 2 1.06 22 | 0.98 
2s SRM Me eme Ne els Pema eean te Weve cecil We (9s. of Sw sue od 13 | 1.06 2 1.04 2% | 1.05 
kes PTW SS ARIE A ai Sim ete I eo ee es 1% | 1.08 2 1208 2 le Le 02 
i) rn Mi eM a Mente Wat oa lauice coo Worse Suet 13 | 1.00 2 0.98 29 led OO 








This table gives arrangement of spiral for which the ratio of volume of spiral 
to volume of concrete core is practically 0.01. These spirals may be used in 
columns designed according to the author’s recommendations on p. 451, and the 
rules of the Cities of Boston, Cleveland, and Philadelphia, given on p. 422. 


930 TABLES AND DIAGRAMS 


Use of Tables 34 to 36.—Tables 34 to 36 give values of piA which are equiva- 
lent areas of vertical steel having same volume as the volume of spiral. They 
were computed from Formula (37), p. 432 and may be used for solving following 
problems: 

1. Find required pitch of spiral, when ratio of spiral, p1, and diameter of core 
are given. In solving, take from the second column of the table the area of core, 
A, corresponding to the given diameter. Multiply this by the given p, to get 
the value of pA. Select tentatively diameter of wire for spiral. In the table 
for the selected size of wire find the pitch corresponding to the diameter of core 
and the value of pA. If the pitch is too small (or too large) use heavier (or 
lighter) wire. 

2. Design Spiral Column by New York Code. The load attributed to the 
spiral in New York Code formulas may be found by multiplying the value pA, 
from the table by 40 000 lb. 

If diameter of column core is known (or assumed) the ratio of vertical steel 
may be accepted and the load carried by concrete and vertical steel found from 
P, = Af.{1 + (n — 1)p] or Af. Table 21 may be used to find f and Diagram 16 
to find P. - If the total load is P, the difference to be resisted by spiral is P — P,. 
Dividing this by 40 000, the value of p,A is obtained. This value located in any 
one of the tables, opposite the diameter of the core, gives the diameter and pitch 
of spiral. The best arrangement of spiral should be selected. 

The problem may also be solved by accepting the diameter of wire and pitch 
of spiral and finding the value of pi1A. This multiplied by 40 000 Ib. gives the 
load resisted by the spiral. To this add the load resisted by the concrete alone 
(area of core multiplied by 600 Ib.). Deduct this sum from the total column load 
and divide by 6 600 lb. This gives the area of vertical steel. | 

3. Design spiral column by Chicago Code. The methods given above also 
may be used with proper modification to design columns according to the Chicago 


Code. 


TABLES AND DIAGRAMS 931 


Table 34.—Values of piA in Spiral Columns. 
Where p; = Ratio of Spiral and A = Area of Conerete Core 
These Values are Equivalent Areas of Vertical Steel having Same Volume as the 
Volume of Spiral 





































3-in. Wire 
Diameter| Area Pitch of 2 Wire. 
of of 
Core Core 

d A Ie 15 2.0 2.25 2.5 2.75 3.0 

10 18.2 2.350 L307 1.72 1.53 1.38 1.26 1.15 
11 95.0 pment spel 00ers 1.68. | 152.) 1 38il, 1.26 
12 113.1 Zal.| 2.00 2.06 1.83 1.65 1.51 1.38 
13 132.7 2990) $en2,56') 2.24 1.99 1.80 1.63 1.49 
14 153.9 sacar gn 2alO-| 2.41 2.14 1.93 1.76 1.61 
15 LiGu7 Seto 29D) 258) 2.290) 2.07 i we 
16 201.1. OEGerie ast t*) 2afOy pe 2.45 | 2.21 2. 1.84 
17 227 .0 SedE 3.35) 2.93, 2.60 |) 2.35 | 2. 1.95 
18 254.5 Pee oO. 00 | OalO Hh 2976.4. 2.49°|) 12:26 |) 2. 07 
19 283 .5 Be ost eo. 2i i aa9l 2.62") 2.39 |. 62.18 
20 314.2 WeGo ih a.0t \o 3-40 9)%3900 1 2.76 | 2.51 2.30 
21 346.4 4.83 4.14} 3.62 3.22 | 2:90 | 2.64) 2.41 
22 380.1 meCOr i 48a )) 8.79 t 8:37 | 3.04 | 2.77:) 2.53 
23 415.5 Peete a O01 8.52, 1 3.18) 2 80-1 2.64 
24 452.4 5.52 | 4.73 4.14| 3.68 | 3.31 3.02 | 2.76 
25 490.9 5.75 aoe 4 OU! bY 3285") 1 3.450) 93.14°) 92-8? 
26 530.9 Ome teste |e 4.48710 O2O8 |7 8,59" ins 27" 1) 2.99 
27 572.6 6.21 5:32 | 4.65 | 4.14 ]..3.73 | 3.39 | 3.10 
28 615.8 6.449 5.52| 4.83 | 4.29 | 3.87] 3.52 | 3.22 
29 660.5 oer teco. (ie) 4.99"|> 4.449) 2 4)00° 138.6516 18.33 
30 706.9 6.90 | 5.91 5.17} 4.60] 4.14] 3.77 | 3.45 
31 754.8 riot) 6211 5.34 | 4.75} 4.28 | 3.89] 3.56 
32 804 .2 7.36 | 6.30 5.51 4.90 | 4.42 | 4.02 | 3.68 
33 855.3 weconte 0.00, |) 6,695 |= 5:06) 42 56.1524215. 1) 13.80 
34 BOTs O gee, 19 6269) )) 5286 |) 5521 4.69 | 4.28} 3.91 
30 962.1 &.00 |r 6.89 6.03 | 5.36] 4.83 | 4.40 | 4.03 
36 1017.9 Brae], 7.09.1) 6.20) |) 5.5205 74.9% |» 4,52 10.414 
37 1075.2 8.51 7 2041) G2a8—| ace Ode teaok hola 4h 65. de 26 
38 1134.1 &. (4) 7.49 |. 6.54 | 5.82 | 95.251. 4.78 | 4.37 
39 1194.6 BO ie 4.68. 1, 6.42 1e.5.987)45.38) |. 4.90 (9.4248 
40 1256.6 9.20 7.88 6.89 |r,6.13 5.52 | 5.03 | 4.60 








Values above upper heavy line are for more than 2 per cent of spiral. 

Values below lower heavy line are for less than 1 per cent of spiral. 

To Find Ratio of Spiral p;.— Divide value from table by area of concrete core. 

To Find Weight of Spiral.—Multiply the corresponding value of p,A by 3.4 
and by the length of spiral plus twice the pitch (to allow for extra turns on 
top and bottom). Add weight of spacers. (See p. 433.) 

To Find Load Carried by Spiral. 

Chicago Code.—Multiply value of p,A from table above by 24nf,., which for 
1: 13 : 3 concrete equals 18 000 Ib. and for 1: 1 : 2 concrete equals 18 125 lb. 

New York Code.—Multiply value of A from table above by 40 000 lb. 


932 TABLES AND DIAGRAMS 


Table 35.—Values of pA in Spiral Columns. 


Where pi = Ratio of Spiral and A = Area of Concrete Core 
These values are equivalent areas of vertical steel having same volume as the 
volume of spiral 


























q,;-in. Wire 
Diameter| Area Pitch of 75 wire. 
oO of 
Core Core 

d A ae) PRES 3.0 
10 18.2 3.14 ere Leon 
11 G5.) 3.45 1.88 iek2 
12 bo 3.76 : 2.06 1.88 
13 or 4.08 oO. ; Deo 2.04 
14 153.9 4.39 oe oe 2.40 220 
15 176.7 Aa. os 2. 25 22300 
16 BOToL 5.02 2. 3: 2.74 Dee 
17 227 .0 5.oo 3) on 2.92 2.66 
18 254 5.65 a aie 3.08 2.82 
19 283.5 5.96 4. on 3.26 2.98 
20 o14.2 Gs20 ae ~A 3.45 3.14 
21 346.4 6.59 4. 43 3.60 3.29 
oe 380.1 6.90 Le 4. aokt 5.40 
23 415.5 ziS BX 4. 3.94 3.60 
24 452.4 fixe oe Lie 4.12 3.76 
25 490.9 ve a Dd. 4.29 3.92 
26 530.9 8. Gx ae. 4.45 4.08 
27 572.6 8.4 6. Bi 4.63 Ae ze 
28 615.8 on 6. 5. 4.80 4.39 
29 660.5 9. 6. 6. 4.97 4.55 
30 706.9 9. vf. 6. 5.14 re a 
31 754.8 9. 79 6. 5.31 | 4.86 
a2 804.2 LO 76. 6. 5.49 5.02 
33 855.3 10. re: 6. 5.66 5.18 
34 907.9 10. ce ve. 5.83 bbe 
a5 962.1 10. Se fh §.00 5.49 
36 1017.9 celle 8.4 “i 6.17 | 5.65 
aire 1075.2 1 8. Te 6.34 5.81 
38 1134.1 Le 8. ; 6.51 5.96 
39 1194.6 12? 9. 6.69 6.11 
40 1256.6 123 9. 6.86 6.28 





Values above upper heavy line are for more than 2 per cent of spiral. 

Values below lower heavy line are for less than 1 per cent of spiral. 

To Find Ratio of Spiral p1.—Divide value from table by area of concrete core. 

Weight of Spiral—Multiply the corresponding value of piA by 3.4 and by 
the length of spiral plus twice the pitch (to allow for extra turns on top and 
bottom). Add weight of spacers. (See p. 433.) 

' Load Carried by Spiral. 

Chicago Code.—Multiply value of piA from table above by 23nfc, which for 
1: 14:3 concrete equals 18 000 lb. and for 1: 1 : 3 concrete, 18 125 Ib. 

New York Code—Multiply value of piA from table above by 40 000 lb. 


TABLES AND DIAGRAMS 933 


Table 36.—Values of p,A in Spiral Columns. 
Where p; = Ratio of Spiral and A = Area of Concrete Core 
These values are equivalent areas of vertical steel having same volume as the 
volume of spiral 

















3-in. Wire 
Diameter; Area Pitch of 4 Wire. 
of of 
Core Core 

d A A EES: 15 . 225 | 25 pares) 320 
10 78.2 4.10 ook 5, Da 2.46 Zeek Nee oe OF 
11 95.0 4.51 3.86 of 3.00 Zeit 2.46 QoDb 
12 Toe! 4.92 4°21 3. o.26 2.95 2.69 2.46 
13 T3a227 DOG Ae DOsinre © BO ey Pea | 2.66 
14 153.9 OTA |. 4:91 4. oLo2 3.44 Del ot 287 
15 L7G 27 6.15 ne 20 4. 4.09 Hess 0 Ov oUsbw o. OF 
16 201.1 Cesar h D262 4. Ae 3.94 SOS) § oLeo 
17. 227.0 CAO he nb OF a 4.64 Daa ers eeell 3.48 
18 2545 foot .0. 32 ie 4.91 4.43 4.03 4-35-69 
19 283.5 TO 667 py B19 4.67 | 4.261] 3.89 
20 Deo ie ae 200). «7.02 Gp l4e15 5.46 4.92 4 48 | 4.10 
21 346.4 8.61 Or 6.45 mee aL dabieeac 14) lene) 
22 380.1 9. re 6.75 6.01 se Bl 4.93 | 4.51 
23 AID. 9. 8. PACOC Oe: a OSOO um eOrto th ol 1 
24 452.4 9. 8. (ead DO Te. On 908. D088. --. 4,92 
25 490 .9 10. 8. TOTS Ose 6.15 DL Out 7O. 12 
26 530.9 10266° | 9213 7.98 TALOUMOROTE tp. 82 bh plod 
27 572.6 11.07 | 9.48 | 8.29 CO Leen 04 OL 05: PMS 03 
28 615.8 | 11.48 | 9.838 8.60 | 7.64 | 6.89 6.27 5.74 
29 660.5 Hoo LOLS. lee 8: 90° 27,92 als 6.50 8 5.94 
30 706.9 12730-10253 9.21 8. fie GEO PG ale 
31 154.8 12-71. 110288 |. 9.52 8. is OF04 a Gaoo 
a2 804.2 Hee 2a th ON 82) he Se (eG TL) 6756 
33 855.3 boot mle bs. LO. 13) fy 9. 8. aoe G2aP 
34 907 .9 13.94 | 11.98 | 10.44 | 9. 8. 7.62 6.97 
BO 962.1 Peewee be LO ica fee 8. Peo Ue ks 
36 1017.9 14-46 | 12.63 | 11.05 0735) e138 805 S06 1° F238 
3d 107522 toot? | 12-98-4011 .386 0-10-1009 104" 8.29. 1-77.59 
38 1134.1 15.58 | 13.34 | 11.66 7 10.37 | 9.35 8.51 7.79 
39 1194.6 15.99 | 13.69 | 11.97 | 10.65 9.59 S24) 799 
40 1256.6 16.40 | 14.04 | 12.28 | 10.92 9.84 | 8.96 | 8.20 





Values above upper heavy line are for more than 2 per cent of spiral. 

Values below lower heavy line are for less than 1 per cent of spiral. 

To Find Ratio of Spiral p:.—Divide value from table by area of concrete core. 

To Find Weight of Spiral—Multiply the corresponding value of p,A by 3.4 
and by the length of spiral plus twice the pitch (to allow for extra turns on 
top and bottom). Add weight of spacers. (See p. 433.) 

To Find Load Carried by Spiral. 

Chicago Code.—Multiply value of »:A from table above by 23n¢, which for 
1: 11 : 3 concrete equals 18 000 lb. and for 1: 1: 2 concrete, 18 125 lb. 

New York Code.—Multiply value of pi1A from table above by 40 000 lb. 


TABLES AND DIAGRAMS 








ETNIES WET 

NT 

NUNN 
il 
i 


\ 
i 
PASTE UTNE NEAT AACE TTT 
nt NININININ 





Hi i Re cat TTT : 

















= 
= 

















4 
tN 


NI 
nth CTUANTAUATNTNIN ATA CTT TTT 


= 
e 


























































































































































































































































































































































































































































































































INIT 
Ht 
COSTA TTA Hl 
TEDSTER AANA NAAT og -o 
AAA N 900 
A NACNDANSNANNAARAANNTI 
NNNINNAANNASNMMNTTIN 
UNNI AAANAATTTT 
N SUMAN ASAT 200 
NINTANANRANNANA Att 
| SAINT 
9 2) S) NS © ly v > \ Le Ss vy ce) ~ © lo 


en JO sen] Df 


(See p. 177.) 


(See p. 177.) 


ants C, for Members Subjected to Direct Compression 
and Flexure. 


Use in Formula for Max. Compression Stress, Te 


D1aGRAM 18.—Const 


935 


RAMS 


1 


q 


AND DIAC 


TABLES 





TINTINT 
TEINTENTEN 


q 








N qn 

NT HHH 
TURININIK 
HATA 
NUUNTINENTENT NTN 





























= 
= 
== 
= 
2 
= 
= 
= 
= 
= 
= 
4 
= 




















ace 
: 
eee 
eS 

eee ee 
= 
= 
: 























<A 
EZ 
a 
ao 
2 
= 
z 
t 4 
= 
Se 
Bes 
= 
2 
a 
=e 
=a 
ee 
: 
a 
eel 


LA 
= 
= 
=o 
S 
2 
ioe 
Se 
i= 
Zz 
ZS 
Pr 
ee 
7 
e 
S Sta: 
+ 











GEST (RR 
Sa= 
S25 
secs 

aes: 
ma P| 
Z| 




















Ze 
va 
eta mee 
Bee 
men 
Ze 
ae = 
aS 
Zoo 
== 
2e= 
mines: 
eee 
Ze 
= 
= 
= 
ee 
= 








= 
= 
=e 
2eeese 
(ees 
= 
Se 




























































































_——— 
2 











= 
= 
Es 
r— 
= 
A 
= 
ZR 
= 
es 
t| 
= 
z= 
= 
2 
+—— 
oe 
S 
ea 
= 
2 
= 
= 
= 


= 
A 
Z| 
[7 
7 
1s 
= 
‘el 
5 GA BF | 
= 
= 
a a a 






























a 

ca 

ma 
2a 
= 

= 

= 

= 

= 
beter 
A P pat 
= 

ZS 

_ 

= 

= 

—= 

= 
Pape 












= 
= 
: 
a 

ze 
su 
= 



























: 
= 
a 
rH 
| 
A 
[7 
ZI 




































































































































































































































































































































































A) & K © ly a ey “I S rr) ® & NN © ey 
WN ~N ~ ~ ~ wn : : ~S ~ 


s 


an fo sano, 


(See p. 177.) 


N 


Use in Formula for Max. Compression Stress f. = Cay. 
(See p. 177.) 


e 
Values of ae 
= 0.9h 


2a 
Flexure. 


Dracram 19.—Constants C, for Members Subjected to Direct Compression 
| and 


TABLES AND DIAGRAMS 















































































































































I 










































































= 
= 
a 
: 
=== 
waa 
See 
== 
Zoe 
ae 
pe A 
ane 




















UNS 































































































7 
we 
z 
4 
= 
— 
A 
= 
= 
= 
Soames 
































q N N \ NN 
CINUTNUINUNTNININININ ACN 
CINTA NTNTNINININTNTNE 


y— 

ZS 
Ss 

74 

















= 
AA 


2 
1 
et 
= 























A 
was 


"iia oie ces cas 





UNIAN TTT 
NAAN. 
NANA NTT TET 


era. 
Bae 
pa Zi 
Bereces 
gee 
te 
==: 
eae 
BEE 
a 
patty 
Ss 
2s 
=== 
— 
= 





a 
Zz 
Gy 
a 
a 
a 
a 
aaa 
= 
ok 
= 
Zz 
ms 
4 
ees) 
= 











NANA HI 
h t\ 


BeBe aeeeee 
Zz Z7 7 = ‘ 
See Bee Sea 
feeeecseos 
ZaaaeoSe 
Sassoon 
eS SS 
gegeesee 
Ses eS Ba eS ee 
2eseesea 
BSSSSSSe 
Sees 
BSS 
Bee eS 2. 
ge2ece= 
geese 
aes oa aes pats 
SSeS = 
= mez Se 
eS ze 


Ze 
2 
Zi A B 
EAA 
Ze 
Be 
ae 
aa 
_ 
A 
= 
Ze 
= 
Zz 
ee 
cS 
5 


TU °° ° 
TTT TTT og 
NUTT 
NN DEUNDATEUAGNONARELL 








NNT ETT 











































































































































































ACH 


























AN 






Values of i 


2a = 0.8h 


(See p. 177.) 


N 
= Car 


Use in Formula for Max. Compression Stresses, f¢ 


C, for Members Subjected to Direct Compression 


DraGRaM 20.—Constants 


(See p. 177.) 


and Flexure. 


937 


TABLES AND DIAGRAMS 


































































































































































































































































































































































































































































































































































































































































































iSUUDDSUUUIDSEUIRN URS SUDSSUUASUD SU SUAS TTASUASS OU U OOM TMU Mm °* ° 
CUT SIN ACSA RA HAVAAVOATELATVGHVGNPOUNOQTOGUIT 
PSST PTT UTTAR Sunn HHUAAUGATAGTOAVEGTVANNGVOQHOGHUGOVSOUNONEINOGHIGUNN 
SE SE Sa aN CS a an A a a a. 
CTU TISUUTSCNSTSAUTNTTTTTNTTN NTN a SSRASRUTTUTTTEECU TUTTE TTEET CCIE CCPC CETTE 
UDNSEUIDS¢ UHDDS€QH SUCRSCUEDAS UMTS UNOS LX RRNA TTT TET TTT 
AUSTEN WG SANSUI CTT 
UPSSECTSSUTSUUTNSSUININ TTR NIST TTT 
PACTS NUTS MUS NUNN TTTCTEEETETEIO 
OTTSSTTTTICUTPTTRUIT TUT TSS SUEUR et 
TTA PSUR BS TNTATIN NINN SASSO AVUNTORVERTONTOQHNI 
PPT BOG NUNS TSSISSSTTTTT ANG 
cite bike DENTIN ARRAS SRR MAL 
UTNE UNIS AVAATUATSUGTOGUNONVRANGQHOQND 
SUPINE 
SUTIN TTT SSS 
hit THREES SSSR RESET 
PSSUTSUTSUTNETNIS IONS TISIRNSSTTTT TTT 
i HUTSON INTUTE 
HECEERAUIEOETUN EA OOHRGSOOEUONERUEAAHaseUUDSSU UM SUaSSUTaSEUDS CU CUCU USN TEE 
ATRATANHNGHNGODGQNUCQNOONOQUESIEQNE HTS RS TSE RNINIEN IRS SSC 
ELECTIVE CNET UTTICETTLCTAPDSOTSSCPSSODS CTS RES ONESTAT 
GAVEATOATNGHUONTOWINATOOH MTT PCTS SSST ESERIES 
HHNCVQOQ00 0NQ000000 00QQQGQQQ0SOQ0Q0U00(S00000000(0HATRATUDIDGSSUDSCRRSSRNSSUNON NUN UOUNNN NU TEE 
SOT TT TTT TT TT TTT TS RARE TONNE USRRAERRENASSAN SUUITTUGAAERIAINI 
HHVEUCROTRBBOEOUUNUHUUOEUREOOSOQTSEOUUOAET REO UOSHMASEOOOOOOOODGSENG EASES UG SMSSNSRNSRN URNS 
FHNNIN0009 SOOSQQQINUAAOCNVONOUON0000Q8000000008000H0RUOOUOOO0Q0QQ001AAIEEOIONDGSSSSSSGS®SSSSSNU 
LANES AUN 
HMA HNVUTHUUUUUAATAPUUUUOUOUUOUUOUNGOO OOOO OQNOOOQOOBOOODDILDGSSSSSEESOSNENSSS NSNSSRNNESNSANSSA AU 
HATEORECUERAOGAYOOUVERRAESOUAO HAN AVAREEOA HOVE AGOOVOVTABOOVUGEAHBOOOAPMHHOBODGSSSSSSSSSISSSSSSSENNSSNSSO TTT 
S dD 9 Nh oO ly tt %y N ~~ =) 3 ey ™ Oo ky 
Nt ~ ~ ~ ~ ~ w ~ ~ ~ ~N 9 S S S S 


99 fo sanjoy 


(See p. 177.) 


bh 


Dracram 21,—Constants C,, for Members Subjected to Direct Compression 


ay 


Ce 


(See p. 177.) 


Values of 
2a = 0.7h 


and Flexure. 


Use in Formula for Max. Compression Stresses f¢ 


938 


100 





TABLES AND DIAGRAMS 








of Eccentricity 
060 050 


d 
040 030 
| 














(ioe SS ES 


020 015 100 




















1] 
Come Hy | _— saa 
Lh $s 


























LN 
an 



























Vj 
y) 
al {Se ae 
[ eneGatie |, ea 
TT 
aS Hii 
Yj 


(fs Ras re 

































































UU AL] 
WY 
ASU 
LTTE] 

















LTT LT 
























































LALA ] 
























































\\N 


\ 





























AN 














mas 

































WNL 


in 


i 
i 
\ 
\ 
\\ 
\ 


NN 


\ 


LZ 





















\ 















\ 
N 
. 
: 











\ 











MI 





i 


AM 





HUNINUVIN 


i 


TWh 


Mt 
\\| 
: 


NS 




















UUM 











| 








i 
\\ 
x 





ia = 








i 














| 
iH 








Hall 
ill 

El 
il 


i 




















Ht 











| 
| 
Nit 


| 
| 














i 
NIN : 
EAN 
inuutgs 
FEEEEEEEEEE TT 

















(=) 











ak 





| 
| 
i 


Ko) 
i 
> 








a 
| 











i 





= 








| 












i 
Hl 






































a 

















30 








20 
Values 


FeO hee 

















15 10 
of Eccentriety " 
h=1.2d. n= 15 








THE 





Values of k, ratio of depth of neutral axis to depth of steel below compressed surface of beam 


As aj ak | 
Both Sides Reinforced. p = bd where A, is Reinforcement at Both Sides. 


DracraMm 22.—Direct Stress and Flexure. Ratio of Depth of Neutral Axis, 


k, for Different Eccentricities. 


Part of Section in Tension. 


(See p. 184.) 


TABLES AND DIAGRAMS 939 













































































































































































0.350 TIV TN 
NAT iil 
\ \ 
vA \' JAA 
ANN i 
Hill 
a) wi 
NIU 
oS Sree NTS 
ah Beaette, esata 
| Mae 
STS 
SIS 9.250 LA aS 
“s ES 
é PTS 
= i Min 4 a 
os i Tegel nist 
a HUT 
S | N TH Hd 
0.200 ee = pe 
7 9220 ATR eet ULI TPE a 
“ sti mee 
© } unas 
< in 
& 0.750 
8 





































































































h=1.2d. n= 15 


A 
Both Sides Reinforced. p = re where A, is Reinforcement at Both Sides. 


DracraM 23.—Direct Stress and Flexure. Constant Cz in Formula (125), p. 182. 
Part of Section in Tension. (See p. 184.) 


940 TABLES AND DIAGRAMS 


Table 37.—Ratio of “ for Different Steel Ratios p. (See p. 186.) 
Steel Ratios p 


~19.002/0.003/0.004/0. 005/0. 006/0. 007|0 .008\0.009)0.010)0.011)0.012 


0.565/0.570/0.576/0.581)/0. 586/0. 590)0 .595|0 . 600/0 .605/0 .609)0.613 








Ne TT el lll! 


a 

| \\ 

cn 
| NN UREA 











0.7 














NACE NN 
NAT a i NIN 


=== 




















wpe? Bi 
Sah SN TAI. > 
col AGASSI & 
s ‘mht a iS AWN : 
S oH NAS SRNTTTTHHTL S 

06 lil HLA WINKS un SR 0.4.0 

: HRN eh RSS|! 

05 nadine te SSS 

STHUAUVUTAUNAUUAAL NaS SHilillins << 

o TMT UTM 

rs i it 

0.2, agit 0.2 


ie ae 
d=09h. n= 15 


As 
Tension Side only Reinforced. p = ao 


DiacraAm 24.—Direct Stress and Flexure. Ratio of Depth of Neutral Axis, k, 
for Different Eccentricities. Part of Section in Tension. (See p. 187.) 


TABLES AND DIAGRAMS 941 



























































































































































































































TL. 

pea TTI ETAT 

tl LL : 

| Ht men 
: HNN ie 
4 °| 
a 0.20°1° 
iN 
|x a.19~ | 
2 lI : 
a NT 0.18 > 
~Im | th Pi 
Ul) 0.171 

3S 

ti ot Clea s 
cya 
tt 0.15 — 
S 
ae 

iM ut : 













































































. 
























































0.3 04 O05 


Values of k 


d=09h. n= 15 
As 
Tension Side only Reinforced. p = bd 
Diacram 25.—Direct Stress and Flexure. Constants C, in Formula (133), p. 187. 
Part of Section in Tension. (See p. 187.) 


942 TABLES AND DIAGRAMS 


Table 38.—Areas, Weights, and Circumferences of Square and Round Bars. 


One cubic foot of steel weighs 490 Ib. 
































| 
b % Square Round bs Square Round 
= rs Bars. Bars. 5 S Bars Bars. 
pail Ae 
Beles Balan | 
pe nes 4 teeipese mlb 5 po a ia 
aa| ". | Ba | se | oo | Be ee] | eee 
as| we “8 a0) ae oh ats) wa a ny =a ik. 
és| ¢ | B2| ¢ | 88] Re Ss] ¢ | Be] Fg | Be | Be 
‘a+ © DH © HO | OF |g+ © ‘Oy o oO | (OR 
= < |e aS) Boe oh) oe 4 |5 = 
rf) 2 | 4.000; 13.60} 3.142) 6.283] 10.68 
7; | 0.004) 0.013) 0.003} 0.196) 0.010 275| 4.254) 14.46) 3.341] 6.480) 11.36 
: 0.016] 0.053] 0.012} 0.393) 0.042|| 24 | 4.516} 15.35] 3.547) 6.676) 12.06 
3; | 0.035} 0.119) 0.028) 0.589) 0.094 23;| 4.785) 16.27| 3.758] 6.872) 12.78 
a 0.063! 0.212) 0.049] 0.785] 0.167|| 2% | 5.063) 17.22) 3.976) 7.069) 13.52 
5 | 0.098] 0.333] 0.077} 0.982) 0.261 275| 5.348] 18.19) 4.200] 7.265) 14.28 
= 0.141] 0.478) 0.110] 1.178] 0.375|| 22 | 5.641] 19.18) 4.430) 7.461) 15.07 
z= | 0.191} 0.651} 0.150) 1.374) 0.511 275| 5.941] 20.20] 4.666] 7.658, 15.86 
S 0.250} 0.850] 0.196] 1.571] 0.667|| 24 | 6.250) 21.25) 4.909] 7.854) 16.69 
2; | 0.316} 1.076) 0.249) 1.767) 0.845 2%;| 6.566] 22.33) 5.157] 8.050) 17.53 
& 0.391] 1.328! 0.307] 1.964] 1.043]| 22 | 6.891) 23.43] 5.412) 8.247) 18.40 
11 | 0.473] 1.608) 0.371} 2.160) 1.262 244) 7.223] 24.56] 5.673] 8.443} 19.29 
3 0.563 1.913] 0.442} 2.356] 1.502]| 22 | 7.563) 25.00) 5.940) 8.639) 20.20 
13 | (0.660) 2.245] 0.519} 2.553) 1.763 213) 7.910] 26.90) 6.213] 8.836; 21.12 
t 0.766! 2.603! 0.601] 2.749] 2.044]) 22 | 8.266] 28.10} 6.492) 9.032) 22.07 
15 | 0.879) 2.989) 0.690) 2.945 2.347|| 248) 8.629) 29.34) 6.777) 9.228) 23.04 
1 | 1.000! 3.400] 0.785] 3.142] 2.670]| 3 | 9.000) 30.60) 7.069} 9.425; 24.03 
13; | 1.129) 3.838) 0.887} 3.338) 3.014 37s| 9.379] 31.89] 7.366) 9.621) 25.04 
13 1.266] 4.3031 0.994! 3.534] 3.379]| 34 | 9.766) 33.20} 7.670) 9.813] 26.08 
133 | 1.410] 4.795} 1.108) 3.731) 3.766 33;|10.160| 34.55} 7.980|10.014) 27.13 
13 1.5631 5.3121 1.227] 3.927} 4.173)) 3% |10.563} 35.92) 8.296/10.210) 28.20 
135 | 1.723) 5.857| 1.353} 4.123) 4.600 375|10.973) 37.31) 8.618)10.407| 29.30 
ie 1.891| 6.428] 1.485] 4.320) 5.049]) 3% [11.391] 38.73) 8.946)10.603) 30.42 
175 | 2.066) 7.026) 1.623} 4.516) 5.518 375|11.816| 40.18) 9.281)10.799) 31.56 
13 2.250] 7.650| 1.767|} 4.712) 6.008]| 34 |12.250) 41.65} 9.621|10.996} 32.71 
1% | 2.441] 8.301] 1.918} 4.909} 6.520 37%5|12:691} 43.14) 9.968}11.192) 33.90 
Le 2.641] 8.978] 2.074) 5.105] 7.051|} 33 |18.141) 44.68/10.321]11.388) 35.09 
134 | 2.848) 9.682] 2.237] 5.301 7.604)| 344/138.598| 46.24/10.680/11.585) 36.31 
12 3 063) 10.41] 2.405] 5.498] 8.178]| 32 |14.063) 47.82)11.045)11.781) 37.56 
123 | 3.285) 11.17] 2.580} 5.694 8.773)| 342/14.535) 49.42)11.416/11.977| 38.81 
1i 3 5161 11.95] 2.761) 5.891] 9.388} 3% |15.016) 51.05)11.793)12.174| 40.10 
128 | 3.754| 12.76} 2.948) 6.087/10.020 343|15.504| 52.71)/12.177|12.370| 41.40 








TABLES AND DIAGRAMS 


Table 39.—Areas of Groups of Bars of Uniform Size. 





of 
Bar in Inches 


Diameter 


[ 


fe sie 


[= 
O\eo | 


Pa 
on 


ale olen 5° ole 


oer ois takes wee 


le lca wl oj: fh 


fox tale 


| 
ale | 


+ 
Sh 


1s oon 5° re 


coo 38 


i 
a 


aie oe 
Olan al alo BO 


cole mI cos pm 


top 


oooo 


ooo °o 





Number of Bars. 


943 








cS 
HS 
oO Oo. 6 


We} 
eo) 
HEH © 


an 
er) 
bee ee 











{2 














31 
.49 
.70 
.96 


.25 
. 58 
.95 
. 36 


81 
. 30 
.83 
.40 


.00) 
.33 


.81 
.45/11.34/13. 


Rounp Bars 


0.29) 0. 


7.36) 8. 
8.91/10. 


.84/10.60|12. 


34 


. 54 
TT 
.05 


37 
74 
15 
.60 


.09 
.63 
21 
.83 


.50 
6.96 


59 


37114. 


Ke OOO 


9): 
39)11. 


. 39 
.61 
.88 
. 20 


.57 
-99 
.46 


NSE 


97 


53 


3 
.15) 4 
.81) 5. 
.52) 6 


.28) 7. 
7.95| 8. 


82/11. 
88/13. 


14}15. 


Square Bars 


bo 

i) 

e 
wWNnNndD = 


9.38/10. 


25/13 .50/15. 


44 
.68 
.98 
34 


15 
~22 
.73 
.31 


94 
62 


.36 
15 


.00 
,86/10. 


94/12. 
23/15. 


75|18. 


se - CO © 


wWwhdo bs 





Ke oO O° 


wWwnNDhND 


tl i) 


Porno re, 


.85 





ll a) 


x - © © 


10. 
12. 


15. 


Ow bd = 


row Nd 


onna 


«J 
J 
eH OO 


aS 
co 
eB who d& 


=] 
=] 
oe kK © 


1Ou) 


.00 


TABLES AND DIAGRAMS 


944 


Table 40.—Area of Square Bars in Slabs for Different Spacing. 


Area of Reinforcement in Square Inches 





wnAOK 
oe TAA HS 
rr sooooo 
lead 2) OR ee Oa Se ne 
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2.18] 2.00] 1.71] 1.50] 1.33] 1.20) 1.00 











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Dia- 
meter 
(In.) 


wo ko (eo mo ojo wo 
is wf e9|0 +2 mice ofS un|oo ae old 2 Pale) “es So 


Di- 
men- 
sion 
(In.) 





Table 41.—Area of Round Bars in Slabs for Different Spacing. 


Area of Reinforcement in Square Inches 





Spacing of Bars. 


12 
In 


9 
In. 
.57| 1.35] 1.18] 1.05] 0.94] 0.78 


0.60} 0.54] 0.50] 0.43] 0.37] 0.33] 0.30} 0.25 
0.74| 0.67] 0.61] 0.53) 0.46] 0.41] 0.37) 0.31 
0.89] 0.81] 0.74] 0.64] 0.56} 0.49) 0.45) 0.37 
1.06] 0.96] 0.88] 0.76] 0.66] 0.59] 0.53) 0.44 
1.24! 1.13] 1.04] 0.89] 0.78] 0.69] 0.62} 0.52 


1.44] 1.31] 1.20] 1.03] 0.90] 0.80] 0.72} 0.60 
1.66] 1.51) 1.38] 1.18] 1.03] 0.92] 0.83] 0.69 


0.47} 0.43] 0.39) 0.34| 0.29] 0.26] 0.24) 0.20 
3.14) 2.69) 2.36] 2.09) 1.88) 1.71] 1 


0.26] 0.24] 0.22] 0.19] 0.17) 0.15) 0.138 


.51] 0.45] 0.40) 0.36] 0.33] 0.30] 0.26) 0.23) 0.20} 0.18} 0.15 


0.18} 0.17} 0.15} 0.13 


5 
In. 
Ced2 


8 
In 





BK 

2 
In. 
0.20 
0.52 
0.66 
0.82 
1,18 
1.38 
1.84 





4 
In. 
0.23 
0.75 
0.92 
1932 
2.07 





1 
2 
In 


0.36] 0.31] 0.26 
0.66} 0.53] 0.44) 0.38) 0.33) 0.29 


1.19) 0.99).0.85 
1.47} 1.23] 1.05 
2.20) 1.78) L482 27) L105 99 


2.65 


3 
; In. ‘ 
PEP AW ML aA OL dle rent 
Sno lie a Ole2eod 
sl 


0.72} 0.60} O 
1.18] 0.94] 0.78] 0.67] 0.59 


1.49 
1.84 


ae 
2 
In 





0.29} 0.23} 0.20] 0.17) 0.15) 0.13 
3.11] 2.48] 2.07] 1.78] 1.56 

3.61] 2.88] 2.40) 2.06] 1.80) 1.60 
4.14 

4.71 


2 
In. 
0.46 
0.90 





wc io wo rio eo wie 
rit anf erico n{ ries ofS s2\c0 TIS cin SIE eleo Sie 


Section 


ees 


Aa 
YI), * 


RECTANGULAR CELL 


}-—— b1- | ae 


4 








HOLLOW CIRCLE | 


For reinforced concrete sections the reinforcement may be considered as replaced by 
having an area equal to n times the area of reinforcement and placed at the same 


(See p. 173.) 


concrete, 


distance from gravity axis. 


* Applicable only to homogeneous (not reinforced) beams. 





TABLES AND DIAGRAMS 


Table 42.—Properties of Sections. 


Moment of 


Area A Inertia I 
bh3 
bh — 
12 
b2 a 
12 
2 Ue 
12 
BH-—bh ds (BH3— bh’) 


Area of flange A1hi2+Asho? 
-+area of web 12 
Ay Ao(hi the)? 
=Ayt4. $O 
7 4(A,+Ae) 
; rt 
ur 7 
a (r4— 114) 
wr%- 17) ars 


Distance of 

Neutral Axis 

from Most 
Strained Fiber * 


h 

%Y 

b 

2 

3bv/2 

H 

a 
ho Athg— Agh 
oo da) 
h Ayjhg— Ach 
to = - --—_———_ 
a 2 2CAy Aa) 

f 

r 


945 


Radius of 
Gyration, 


Nf iA 





0.289d 


0.2896 


0.2896 


| BH3— bh3 


‘12(BH—bh; 





wv 





INDEX 


A 


Abrams, D. A., 52, 53, 55, 57 
Acme flat slab system, 362 
. Adjustable metal tiles, 605 
Aggregates, choice of, 6 
coarse, 3 
coloring surface by, 737 
effect on modulus of elasticity, 201 
fine, 2 
floor surfaces, 621 
grading, 3 
selection, 1 
size, 273 
Air space below roof, 652 
Alden Park Manor, 748 
Anchored ends, working stresses, 263 
Anchoring bars, 265 
chimneys, etc., 477 
column reinforcement, 269 
compression reinforcement, 268 
footing reinforcement, 537 
hooks, 163, 269 
plate, 163 
problems of, 267 
retaining wall reinforcement, 853 
steel in wall columns, 317, 462 
straight imbedment, 163 
web reinforcement, 244 
Angle of bent reinforcement, 291 
column heads for flat slabs, 320 
Apartment de Luxe, metal tiles for, 607 
houses, 805 
Arch roof, details, 676 
Arches. See also Vol. III. 
fulerum girders as, 772 
Area, column sections, 915 
reinforcing bars, Table 12, 942, 943, 
944 


Armory at Danville, 673 

Assembling column steel, 418 

Assumptions, rectangular beam theory, 
126 

Auditoriums, concrete, 762, 810 

Automobiles, dimensions of, 796 

Auxiliary structures above roof, 678 

Axial compression and bending, col- 
umns, 463 


B 
Bach, C., 24, 38, 49, 50, 52, 58, 59, 60, 
62, 65, 93, 95 
Backing of curtain walls, 745 
Balanced design, 203 
rectangular beams, 132 
Balcony, cantilevers, 772 
cantilever trusses, 778, 782 
design, 763 
floors, 778 
framing, 780 
fulerum girder for, 765 
Barrett’s Specifications, tar and gravel 
roofs, 655 
Bars. See also Reinforcement 
aluminum identification tags, 723 
spacers, 211 
table of sizes, 12, 942, 943, 944 
Barton spider-web reinforcement, 359 
Base for structural steel columns, 443 
Basement slabs, 618 
supported by columns, 619 
thickness, 619 
under water pressure, 619 
Basement walls, 637, 638 
waterproofing of, 637 
Batter, retaining walls, 841 
Beam, design tables, 880 


947 


948 


Beam and girder schedules, 716 
and girder type floors, 575 
Beam and slab, 383 
foundation, 540 
Beams. See also Wall beams 
See also T-beams 
balanced design, 1382, 203 
bending moments to use, 275 
compression below flange, 137 
constants for rectangular, 205, 880, 
881 
continuous at support, 281 
depth, formula, rectangular, 131, 204 
design constants, 205 
details on plans, 723 
diagonal tension in, 147 
formulas for, 247 
reinforcement for, 248, 252 
tests, 38 . 
failure by compression, 50 
by tension, 35 
false, 615 
formulas, 
design, rectangular beams, 204 
steel top and bottom, 234 
T-beam, 216 
unsymmetrical T-beams, 142 
wedge-shaped beams, 140 
review, rectangular beams, 207 
T-beams, 224 
haunch, 284, 405 
hollow circular, 822 
neutral axis, rectangular, 130, 207 
steel top and bottom, 139 
T-beam, 134, 220 
numbering, 711 
reinforcing for, 41 
resisting forces, 129 
restraint of, 275 
in flat slab, 377 
safe loads, table, 884 
schedules for, 715 
shearing stresses, 143, 150 
spacing of tension bars, 273 
spandrel, 334, 459, 746 
steel in top and bottom, theory, 137 
simplified formulas, 234 
diagrams, 904 to 909 


INDEX 


Beams, straight-line theory, 127 
and actual conditions, 142 
stresses, actual and computed, 24 
test of continuous, 64 
rectangular, 20 
steel top and bottom, 48 
T-beam, 35 

theory of rectangular beams, 124 
steel top and bottom, 137. 
T-beam, 133 
unsymmetrical T-beams, 142 
wedge-shaped beams, 140 

unbalanced design, 132 

wall, 379, 578 

width required by reinforcement, 

Table, 274 


-Bearing stresses, 270 


brickwork, 631 
Bend, angle of reinforcement, 291-297 
test, 8 
Bending, columns subjected to, 458 
combined with, direct tension, 189 
direct compression, 187 
thrust, 187 
Bending bars, point of, 287 
slabs, 355 
beam reinforcement, points of, 292- 
297 
diagrams, for beam reinforcement, 
292-297 
Bending moments, beam design, 275 
building frames, 279 
coefficients, from test of continuous 
beams, 65 
combined footings, 524 
continuous beams, 277 
diagrams for determining point of 
bending of bars, 292-297 
exterior columns, 314 
flat slabs, 328 
See also Flat Slabs 
interior panels, 331 
wall panels, 333 
footings, 495 
permissible variations in flat slabs, 
335° 
strap beams, 533 — 


Bending sketches, 724 


INDEX 


Bent bars, marking of, 724 
Bent-up bars as web reinforcement: 
area and spacing, 156, 249 
example, 582, 587 
formulas for, 156, 249 
limiting spacing, 250 
Bergstrom, Edwin, 764, 779 
Bignell pre-cast pile, 552 
Bond, allowable unit stresses, 263 
anchorage in beam reinforcement, 
effect on, 26 
deformed bars, 53, 263 
design formulas for, 262, 268 
distribution of stresses, 57 
surface of bar, 53 
on temperature cracks, 299 
elements of, 51 
function of external shear, 261 
green concrete, 56 
hook in beam, effect on, 263 
influence of age, 54 
of mix, 54 
length of lap for, table, 414 
load, effect of position on, 58 
mix, effect of, 59 
necessary to develop tensile strength 
of bar, 354 
plain bars, 53, 263 
problems of, 264 
proportions of concrete, effect, 59 
ratio length of imbedment to diam- 
eter bar, 52 
resistance, 52, 260 
shape of bar, effect, 53 
steel to concrete, explanation, 260 
stirrups, 40 
strength, 50 
stress, requirements, 132 
stresses, 10, 56, 151, 263 
flat slabs, 351 
footings, 499, 500, 528 
fulerum girders, 768 
importance of computation, 261 
small vs. large bars, 262 
unit stresses, allowable, 268 
Bonded finish for floors, 621 
Borings, results of, 469 
Boston Building Code, 347 


949 


Boston Building Code, bearing values 
of soils, 471 
columns, 408, 421, 488, 448, 450 
live loads, 454, 570 
structural steel and 
umns, 438 
width of flange in T-beam, 218 


conerete  col- 


' Bracket loads on columns, 464 


Bransom System, floors, 588 
Brick, exterior, 747 
Brick parapets, 658 

piers, strength, 77 

stresses on, 78, 631 

veneer building, 749 

wall and concrete frame, 739 

wall thickness, 31 
Brickwork, bearing walls, 335, 629 
Bridges. See Vol. III 
British building rules, 202 
Brush, finish, 736 

hammer finish, 737 
Buckling. of columns, 81, 87 
Building Codes. See Code in Ques- 

tion 

Building construction, 564 
Building, frames, formulas, 279 

lines, 721 

plans, 707 

tests on completed, 113 
Bunte Bros., factory, 681 
Bureau of Standards, 77 
Butting bars, 415 
Buttresses, retaining walls, 833 


ae: 
Ca, values of, 939 

Cp, values of, 941 

Ca, values of, 894, 895 

C., values of, 934 

Cn, for square footings, 495 


_ Cp, for rectangular footings, 495 


Cy3, for square footings, 497 

Cys, for rectangular footing, 498 
C;;, for rectangular footing, 502 
Cp, values for earth pressure, $37 
Cr, table of values, 896 

Ci to Cy, diagram, 911 


950 


C0 and Ci, table, 912 
C12, table, 913 
C13, table, 914 
Caisson pile, 560 
Caissons, 477 
Cantilever balcony truss, Grauman’s 
Theater, 779 
vs. ordinary beam, 775 
Cantilever retaining walls, 849 
Cantilevers, design of, 772 
Capacity of stairways, 684 
Carrying capacity, soil, 469 
wood piles, 542 
Carter’s Ink Co. Factory, 734 
Cast-in-place piles, 554, 558, 559 
Cast-iron core, concrete columns, 445 
Cast stone, 733 
Ceilings, ornamental, 614 
suspended, 19, 613 
Cellular walls, 869, 871 
Cement, 2 
flash set, 2 
mixing, 4 
protection of reinforcement, 9 
ramming, 2 
slump test, 4 
testing, 2 
water in, 4 
Chairs, metal, 356 
Cheapest exterior treatment, 730 
Chenoweth pre-cast pile, 551 
Chicago Building Code, 218, 322, 347 
columns, 396, 408, 421, 425, 439, 
448, 451 
exits, 683 
flat slabs, 312, 317, 396 
live loads, 454 
Chimney, Edison Electric Ill. Co., 
Brooklyn, 824 
Chimneys, concrete, 812 
constants for neutral axis, 828 
construction, 827 
design, 822, 827 
essentials, 827 
mix of concrete for, 823 
temperatures, 822 
thickness concrete, 819 
Cinder fill, roofs, 653 


INDEX 


Cinder fill, foundations, 540 
Circular beams, 822 
plain sections, stresses in, 172 
reinforced section, 179 
Circular ramps, 801 
Circumferential flat slab  reinforce- 
ment, 117, 367 
Clark Biscuit Co. Building, 793 
Clearances, elevator shaft, 696 
Cleveland Building Code, 218 
columns, 408, 438, 448, 450, 464 
Coefficient of contraction, concrete, 300 
Cold weather, laying in, 299 


| Cold-storage, live loads, 792 


warehouses, 792 
Coloring of surfaces, 737 
Columns. See also Exterior Columns. 
See also Interior Columns. 
See also Spiral Columns. 
See also Wall Columns 
Column caps, 760 
Columns, 403 
action under test, 86 
allowable unit stresses, 
axial compression and bending, 463 
composite columns, 436 
plain concrete columns, 879 
spiral columns, 421 
vertical bars only, 406 
axial compression and bending, 463 
bars, 412 
bars, lapping of, 413 
butting, 415 
in spiral column, 434 
bearing stresses, 271 
bending in, 458 
bracket loads, 464 
buckling, 81, 87 
building codes for, 408, 409, 422, 
437, 450 
cast-iron core, 445 
comparison of cost, 445 
composite, 436 
concrete, effect of richness, 76 
concrete vs. brick, 77 
core. See Effective area 
cost of concrete vs. forms, 445 
crane loads, 464 


INDEX 


951 


Columns, deformations, plain steel, 89 | Columns, reinforcement, influence of, 80 


design example, 456 
dowels, 417 
eccentric load on, 164 
economy of design, 445 
economy spiral vs. vertical steel only, 
449 
effective area, 406 
exterior. See Exterior columns 
factor of safety, 79 
failure, 76, 79 
fireproofing of, 272, 407 
flat slab construction, 305 
formulas, general, 158 
for design, long columns, 435 
plain concrete, 404 
spiral columns, 421, 422, 424, 425, 
426 
vertical steel only, 406 
heads for flat slab, 319 
angle of, 319; 320 
slab thickness at, 338 
stresses at, 199 
hinged effect at bottom, 668 
interior for flat slabs, 305, 310, 312 
lapping of reinforcement, 413 
limitation of size for flat slabs, 307 
limiting length, 407 
live load on, reduction of, 453 
load on footings, 479 
long, formulas for, 434 
tests on, 93 
mix, influence of, 80 
modulus of elasticity, 76 
’ numbering, on floor plan, 710 
oblong spiral, 431 
octagonal spiral, 429 
octagonal vs. round, 429 
orchestra floor, 763 
plain, test of, 81 
point of maximum bending moment, 
459 
Poisson’s ratio, 76 
proportions of concrete for, 446 
ratio of rigidity in flat slab, 459 
reduction of live load, 452 


reinforcement, anchorage of, 269, 413 | 


disposition of, 414 


uniform size of, 412 
repeated loads, test under, 77 
review of design, 411 
round with 1 per cent spiral, tables 
of loads, 926, 927, 928 
round with vertical reinforcement, 
tables of loads, 923, 924, 925 
schedule for, 719 
spacing in garages, 797 
spiral design of, 419, 421, 422, 430, 
448 
test of, 82, 89 
splicing structural steel, 441 
splitting effect of bars, 79 
square columns, table, 920, 921, 922 
square spiral not effective, 430 
test of, 88 
steel, assembling of, 418 
strength, formula for, 160, 421 
stresses in spiral, 422 
with vertical steel only, 205 
structural steel in concrete, 435 
supporting basement, 619 
tensile steel, amount, 461 
test of, 76 
ties, 417 
total loads on, 919 
unsupported length, 405 
‘vertical and spiral reinforcement, 
_ test of, 84 
vertically reinforced, test of, 78 
yielding of, 108 
Combined footings, 522 
bending moments, 524 
external shear, 526 
for interior columns, 523 
shape of base, 522 
unequal loads on, 522 
Comparative costs of columns, 445 
floors, 786 
Comparison of floor surfaces, 627 
Composite columns, 436 
piles, 555 
Compression, below flange in T-beam, 
137 
due to eccentricity, 191 
failure of beam by, 30, 32 


952 


Compression, reinforcement for beams, 
50 
anchorage of, 268 
diagrammatic solution, 237 
formulas for design, 234 
flat slabs, 345 
influence on deflection, 63 
problems in design, 232 
ratio of, 236, 283 
supports, 282, 283 
T-beams, 225, 230 
stress in concrete at 
223 
stresses in flat slabs, 341 
Concentrated loading, stirrups for, 254 
Concrete, definition of. See item in 
question 
Concrete buildings, cost, 567 
height, 803 
shape of, 804 
types, 572 
Concrete chimneys, 812 
coefficient of contraction, 300 
Condensation under roofs, 649 
Condron Company, 362 
Considére, 22 
Consistency of concrete, 26 
Constants. See also C4, ete. 
beam and slab design, 205, 880 
Continuous beams: 
design, 285 
details, 281 
diagrams for points of bending 
bars in, 293-297 
equal span, formulas for, 278 
minimum depth, 222 
points of inflection, 289 
span of, 277 
tension in steel at supports, 288 
tests, 64, 115 
unequal spans, 280 
unsymmetrical loading, 280 
vs. building frames, 277 
web reinforcement for, 157 
Continuous slabs, formulas for, 209, 
279 
arrangement of reinforcement, 210 
wall column footings, 518 


supports, 


INDEX 


Coping, 658 
Corner column footings, 514 
Corrugated Bar Co., 99, 105, 363 
Cost, columns, 445 
columns and footings, 577 
reinforced concrete buildings, 567 
spiral reinforcement, 434 
various floors, 786 
types of construction, 564 
Counterforts, 833, 863, 867 
Coverings for roofs, 654 
Cracks, 20, 66 
cantilever slab, 119 
diagonal tension, 40, 147, 153, 157 
percentage reinforcement, 300 
shrinkage, 300 
spacing of, 300 
temperature, 299 
Crane loads on columns, 464 
Cret, Paul P., 629 
Cross bars, 37 
fulcrum girders, 768 
in slabs, 211 
over girders, 211 
Cross-section Massachusetts Inst. Tech., 
808 
theater auditorium, 781 
Cubes, strength of, 50 
Cummings pre-cast pile, 552 
Curb bars, 617 
Curbs for garage drives, 798 


Curtain wall backing, 745 


Curtain walls, 572 


D 


Dash coat finish, 736 

Dead load, balcony cantilevers, 774 
on floors, 571 
roofs, 648 é 

Deere and Webber Building, test on, 

113 

Deflection, beams, 61, 115 : 
compressive steel, influence on, 63 
end beams and wall columns, 460 
flange width, influence on, 62 
slabs, 69, 103, 198, 326 
‘steel, influence of, 61 
theoretical and actual, 65 


INDEX 


Deformations, concrete, 72, 341 
in reinforcement, 28 
of columns, plain steel, 89 
plastic in slab, 109 
relations to stress, 387 
spiral colums, 82 
steel, 72, 117 
‘Deformed bars, purpose of, 10 
Delaware, Lackawanna and Western 
R. R., 834 
Densmore, Le Clear and Robbins, 733, 
734, 740, 752, 759 
Depth of beam increase by haunch, 284 
beams, formula for, 131 
cantilevers, 775 
footings, 493 
foundation, 470 
Designation of structural members on 
plans, 710 
Detail of column cap, 760 
Detailing methods, 716 
D’Hume Ramp System for garages, 801 
Diagram, solving stress by, 176 
Diagonal tension: 
allowable unit stress, 244 
assumptions for, 153 
beam and tile floors, 596 
bent bars for, 248, 251 
chimneys, 819 
combined footings, 527 
distribution of between concrete and 
steel, 245, 246, 247 
element of beam under, 152 
failure by, 46, 147 
footings, 482, 502 
for moving loads, 258 
formulas for, 148, 247 
reinforcement in flat slab for, 350 
reinforcement for, 247 
requirements for web design, 218 
stirrups for, 247, 252 
stresses, 241 
test, 35, 38, 48 
uniformly distributed load, 156 
Diamond drill borings, 469 
Direct stress and bending, 169. 
See also Thrust and Bending mo- 
ment 


953 


Direct stress and bending, allowable 
stress, 463 
circular section, 179 
formulas for design: 
total section in compression, 177 
part section in tension, 180, 182, 
186 
negative thrust and moment, 167, 
189 
plain section, 169 
reinforced section, 173 
sign of bending moment and position 
of thrust, 166 
steel in tensile face only, 189 
tension, 162 
Distributing beams, 633 
Dowels, for columns, 417 
on plans, 720 
pedestals, 487 
retaining walls, 863 
Drainage, retaining walls, 834 
roofs, 649 
Drop panels for flat slabs, 322 
advisability of, 322 
at wall columns, 324 
dimensions, 323 
shear stress at, 914 


1) 


Earth, weight of, 833 
Earth, pressure, basement walls, 638, 
644 
distribution of, 835 
passive, 839 
Rankine’s theory, 835 
retaining walls, 841, 842 
special cases, 836 
surcharge, 837 
variation of, 640 
Eastern Concrete Construction Co., 60 
Eccentric footings, 476 
Eccentric loading, 164 
See Direct Stress and Bending 
See Thrust and Bending Moment 
Eccentric thrust, relation to bending 
moment, 165 
stress caused by, 171 


954 


Eccentricity, effect of, 177 
limiting, 178 
Economical, arrangement floor beams, 
575 
depth, T-beams, 220 
design, column reinforcement, 447 
rectangular beams, 213 
mix for columns, 447 


Economy, cantilever vs. counterfort 
walls, 863 
columns, spiral vs. vertical steel only, 
449 


concrete construction, 564 
concrete vs. rubble masonry, 832 
deformed bars, 10 
floor construction, 806 
foundation depth, 470 
plain vs. reinforced walls, 832 
reinforced concrete, 123 
reinforced concrete footings, 481 
reinforcing steel, 9 
rich mix, 213 
structural steel 
columns, 452 
T-beams vs. truss, 772 
various types floor, 785 
Edison Electric Illuminating Co. Chim- 
ney, 824 
Effective area, columns, 406 
section for thrust and bending, 179 
width slabs, 72 
Elastic flashing, 659 
joints, ground slabs, 618 
limit, 50 
Eleannes reinforcing bars, 14 
Elevator pit, 696 
provision for additional, 791 
shafts, 682, 695 
Elevators, warehouse, 791 
Eliot St. Garage, Boston, 801 
Emperger, 93 
Engineering News pile formula, 543, 
559 
Exits required in buildings, 683 
Expanded metal reinforcement, 15, 18 
Expansion joints in roofs, 653 
Exterior columns, bending moment 
on, 314 


and concrete 


INDEX 


Exterior columns, design of, 312, 460 
rigidity, 313 
Exterior treatment buildings, 728 


F 


fc, allowable unit stress, 879. 
fs, allowable unit stress, 879 
Faber, Oscar, 812 
Facing with terra cotta, 753 
Factor of safety, 30 
columns, 79, 80 
definition of, 125 
reinforcing steel, 8 
retaining walls, 842 
Factory, 
floors, 795 
Failure, at footing, 479 
cantilever walls, 850 
green concrete, 46 
retaining walls, 841 
torsion, 95 
wall with counterforts, 865 
False beams, 615 
Fillets for T-beams, 146 
Finish, brush, 736 
exterior, 728 
hammer, 737 
interior, 755 
floor surfaces, 620 
Fire exits, 682 
protection, reinforcement in beams, 
272 
Fireproof partitions, concrete, 635 
Fireproofed columns, 88, 93, 407 
Fireproofing, 422 
First baleony fulcrum girder, 783 
Flange, compression below, 137 
T-beam, compression neglected, 133 
T-beams, 36 
width, influence deflection, 62 
Flanges, shearing in, 37 
Flash set of cement, 2 
Flashing, 659 
Flat ceiling, 610 
slab design, 911 
Flat slab, action of, 198 
action of rings, 373 


INDEX 


Flat slab, advantages of, 304 
beams, 377 
bending moments, 328, 331, 332, 334 
Chicago Code, 398 
New York Code, 399 
bond stresses, 351 
Chicago regulation, 396 
columns for, 305 
bending moments in exterior, 314 
exterior, limitation of, 312 
interior, limitation of, 307 
column heads, 319 
author’s rules, 319 
Chicago Code, 397 
New York Code, 397 
compression at column head, 342 
in center, 346 
Chicago Code, 401 
New York Code, 401 
compression reinforcement, 345 
description of, 304 
design details, 360 
design strips, 330 
diagonal tension reinforcement, 350 
drop panels, 322 
authors’ rules, 322 
Chicago Code, 397 
New York Code, 397 
example of design, 390 
formulas for area of steel, 351 
formulas for bending moments, 331 
interior panel, 332 
wall panels, 334 
formulas for compression stresses, 
343 
at column head, 344 
in center, 346 
interior columns, 305 
authors’ recommendation, 307 
Chicago Code, 396 
New York Code, 396 
maximum bending moment, 
tribution of, 330 
method of design, 389 
New York regulation, 397 
openings, in, 375 
points of inflection, 199 
problems in design, 389 


dis- 


955 


Flat slab, sections of maximum bend- 
ing moments, 329 
separated in simple parts, 371 
shearing stresses, 346 
critical sections for, 348 
formulas for, 349 
Smulski system, 369, 376 
test of, 104 
steel, working unit stresses, 351 
Chicago Code, 401 
New York Code, 401 
stresses at column head, 199 
stresses in central portion, 200 
suspended action impossible, 387 
systems: 
Acme system, 363 
Barton’s spider web system, 359 
four-way system, 358 
Smulski system, 367 
three-way system, 374 
two-way system, 362 
theory, 194 
thickness of slab, 325 
authors’ rules, 327 
Chicago Code, 397 
New York Code, 397 
thickness of slab at column head, 
formula, 338 
thickness of clab in center, formula, 
340 
two-way system, 364, 365 
test of, 99 
wall beams, 379 
above slab, 381 
below slab, 379 
wall columns, 312 
authors’ recommendation, 312 
Chicago Code, 397 
New York Code, 397 
Flexure, 164, 173 
Floor, girders supporting joists, 597 
joists, 592 
loads, 569 
plans, 708, 713, 722, 804 
surfaces, 620, 627 
Floors, balcony, 778 
beam and girder design, 575 
example, 578 


956 


INDEX 


Floors, concentrated loads on light- | Footings, diagonal tension in, 502 


weight, 609 
concrete and tile, 597, 689 
construction of light-weight, 609 
cracking of topping, 594 
dead loads on, 571 
economical arrangement of beams, 
575 
economy, relative of different types, 
786 
factory, 795 
finish, 805, 806 
flat slab construction. 
slab 
garages, 795 
granolithic finish for, 622 
hardwood finish for, 625 
hollow tile, 590 
hospitals, 807 
light-weight, 588 
painting of, 623 
school buildings, 808 
steel tile, 602 
structural steel and concrete, 613 
tile, 590 
two-way tiling, 592 
types, 574, 787 
warehouse, 791 
wood block, 626 
wood finish, 624 
Footing plans, 721 
Footings, 469, 481 
See also combined footings 
area of, 473 
bending moments on, 495 
bond stresses in, 499 
column loads on, 479 
combined, 522 
concentric, 476 
constants for formulas, 495, 497, 498, 
502, 515 
continuous wall column, 518 
corner column, 514 
critical sections of plain concrete, 481 
of reinforced footing, 496 
for diagonal tension, 502 
dead loads on, 536 
depth of plain table, 481 


See Flat 


eccentric, 476 
external shear in, 500 
failure of, 479 
independent column, 485 
independent stepped, 490 
piles under, 516 
plain concrete, 480 
projection of wall, 520 
proportions of independent, 491 
punching ‘shear on, 492 
rectangular, 510 
schedules, 721 
sloped, 484, 491, 506 
sloped vs. stepped, 491 
stepped, 484, 487, 490, 506 
strap beams for, 533, 536 
stresses, variation of, in independent, 
489 
types, 481 
wall, 482 
Formulas. See subject 
notation for, 128 
Foundations, 469 
at property line, 477 
beam and slab, 540 
design loads for, 478 
encroachment on public streets, 478 
flat slab, 539 
outside columns, 478 
plans, 712, 721 
plates for, 417 
raft, 538 
steel cylinder, 562 
Foundry for Garfield, Smith & Co., 666 
Four-way flat slab details, 718 
system reinforcement, 358 
Framing plan, 714 
Frank E. Davis Fish Co., building, 
733 
Franklin Service Station, 802 
Freytag, 87, 88 
Frictional resistance of piles, 543 
Frost, power of, 835 
Fulcrum girders for 
769 
Furring, 614 
inside walls, 746 


theaters, 765, 


INDEX 


G 

Garages, 795 

column spacing, 797 

inclined driveway, 799 

live loads, 796 
Garfield & Smith Co., 665 
German building rules, 202 
Giant pre-cast pile, 552 
Girders, numbering on plans, 711 

schedules, 715 

supporting joists, 597 
Glass, sizes of, 705 
Glazing, 703 ' 
Glenday, Charles, 812 
Goldbeck, A. T., 69 
Goosenecking, 415 
Gow, Charles R., 547 
Graf, O., 95 
Granolithic floor finish, 609, 620, 623 
Graphical spacing of stirrups, 252 
Grauman’s Theater, 671, 674, 764, 779 
Gravell, William H., 629 
Gravity walls, design of, 839 
Gymnasiums, 810 
Gypsum tiles, 592 


H 


Hammered surface, 738 
Hardwood floors, 625 
Hatt, W. K., 99, 104, 112 
Haunches, 284, 405 
Havemeyer bars, 13 
Heel cantilever walls, 852 
Height of concrete buildings, 803 
Herringbone lath, 611 
Hickey for bending bars, 355, 415, 727 
Hide and Leather Building, 738, 804 
Higginson, Wm., 681 
Hinged effect, bottom of columns, 668 
Hollow circular beams, 822 
tile, 589, 590, 598, 809 
Homogeneous beams, stresses-in, 241 
Hooks in reinforcement, 59, 60, 63, 266, 
269 
Hoops. See Spiral reinforcement 
chimney reinforcement, 819, 825 
strength, 820 





957 


Hospitals, 805 
design, 808 
State of Wisconsin, 807, 808 
Hotels, 805 
Howard, J. E., 80 
Hughes, C. A., 97 
Humidity, 652 
Humphrey, Richard L., 25, 27, 62 
Hy-rib, 19, 611 


I 


Ideal Shoe Factory, 793 
Identification tags on bars, 723 
Imbedment, formulas for, 267 
length, 267, 462 
Imitation marble, 760 
Inclined drive, garage, 799 
stirrups, 243 
Independent footings with piles, 516 
Inertia, moment of, 174, 281 
Inflection, coefficient of, 65 
points of, 110, 289 
Inserts for shafting, 615 
Insulation of roofs, 650 
Interior, cclumns, 305, 310, 312 
finish, 755 
office building, 761 
panels, bending moment in, 331 


J 


j, formulas, rectangular beams and 
slabs, 130, 207 
steel top and bottom, 240 
T-beam, 134, 221 
j, rectangular beams and slabs, tables, 
205, 880, 881 
j, steel top and bottom, diagram, 910 
4, T-beam, table, 221, 897 
Jetting, pre-cast piles, 551 
Joint Committee, year 1916, 246, 448 
year 1924, 161, 245, 246, 282, 
324, 331, 333, 337, 405 
1924 Full Report. See Vol. IT 
Joist, depth, 596 
loads from slabs, 74 
Joists for floors, 594 


958 


K 


k, diagrams, direct stress and flexure, 
938, 940 
rectangular beams, 883 


k, formula, direct stress and flexure, 


183, 198 
rectangular beam, 130, 207 
steel in top and bottom, 139 
T-beam, 134, 137 
k, tables rectangular beams and slabs, 
205, 880, 881 
T-beam, table, 897 
Kahn, Arthur, 803 
Kahn trussed reinforcing bars, 13 
Kellog, Harold Field, 748 
Kenneth, M. de Vose and Co., 748 
Klein, Arthur F., 785 


L 


L-shaped retaining walls, 849 
Lacher design, walls, 872 
Laitance, 5 
Lap, plan bar, 121 
Lapping of bars, length, 414 
Life Savers Inc. Building, 743 
Limestone, 272 
Lindenthal, Gustave, 834, 873 
Linoleum floor finish, 805 
Lintel beam, 103 
Live load, balcony cantilevers, 774 
beams, table of, 884 
building codes, 454 
buildings, 452, 570, 805 
cold storage, 792 
columns, 453 
floors, 569 
garages, 797 
reduction for columns, 452 
authors’ recommendation, 453 
Boston Code, 454 
Chicago Code, 454 
New York Code, 454 
reduction for large areas, 454 
school floors, 808 
Load transference, flat slabs, 386 
Loading, concrete beams, 20, 884 


INDEX 


Loading, concrete chimneys, 812 
eccentric, 164 
effect on neutral axis, 27 
effect on reinforcement, 27 
floors, 569 
for foundation design, 478 
office floors, 805 
platforms, 790 
roofs, 648 
slabs, 69, 102, 886 
slab stress at maximum, 109 
slabs to joists, 74 
tests of buildings under, 113 
Lockwood, Greene Company, 743 
Long columns, 434 
Long column span roof construction, 
661, 671 
Longitudinal reinforcement. See also 
Vertical reinforcement. 
for shear, 146 
fulcrum girders, 768 
steel, columns with, 405 
Lord, A. R., 88, 113 
Losse, L. H., 25,27, 62 


M 


MacMillan, A. B., 626, 627 
Manufacturing buildings, 792 
Marking of bent bars, 724 
Mass. Institute of Tech., facades, 754 
framing plan, 714 
section, 808, 810 
Maurer, EK. R., 220 
Medical Arts Building, Dallas, 803 
Melan arch, 122 . 
Mensch, L. J., 671 
Metal flashing, 659 
lath, 18 
roofing, 656 
Mills, Charles M., 299 
Minimum depth T-beam, 219 
Mix of concrete, basement slabs, 619 
concrete chimneys, 823 
concrete columns, economy of, 447 
for use, 4 
fulcrum girders, 772 
granolithic finish, 620 


INDEX 


Mix of concrete, influence of, 24, 80, 
83, 87 
pre-cast piles, 550 
stress in steel and concrete, due to, 
87 
Modulus of elasticity, 31, 80 
columns, 76 
concrete, 38, 202 
effect of, if constant, 127 
for beams, 127 
variation with stresses, 142 
rupture, plain concrete beam, 48 
Moisture protection of steel, 273 
Moment arm, formulas, beam with 
compression steel, 240 
rectangular beams, 130, 207 
T-beam, 134, 221 
tables and diagrams, beams with 
compression steel, 910 
rectangular beams, 205, 880, 881, 
883 
T-beams, 897 
Moment coefficients, 102 
Moment of inertia, effect on bending 
moment, 281 
for different sections, table, 945 
Moment of resistance, rectangular 
beam, 130 
T-beam, 136 — 
Monotype reinforcing bars, 14 
Morris, C. T., 70, 74 
Morrow, D. C., 374 
Moving loads, diagona: tension for, 258 
Mullions, 699 
Muntins, 699 
Mushroom system of reinforcement, 


359 
N- 
National Board of Fire Underwriters, 
685 


National Fireproofing Company, 590 
Negative thrust, 165 
Neutral axis, formulas, direct stress and 
flexure, 183, 193 
rectangular beam, 130, 207 
steel in top and bottom, 139 


959 


Neutral axis, formulas, T-beam, 134, 
137, 224 
tables and diagrams, direct stress 
and flexure, 938, 940 
rectangular beams, 205, 880,881, 
883 
T-beam, 897 
tests, 27, 35 
New Channon Building, test, 342 
New York Building Code, 160, 218, 322 
ender concrete, 612 
columns, 396, 408, 421, 424, 437, 448 
450, 464 
flat slab regulations, 347, 396 
live loads, 454 
panels, 399, 400 
piles, 544 
smoke-proof stair towers, 686 
spiral columns, 426, 918 
stairways, 685 
steel cylinder foundation, 562 
New York Connecting R. R., 873 
Nichols, John R., 196 
Nielson, N. S8., 196 
Norcross system of reinforcement, 358 
Northup, W. C., 784 
Notation, 128 
Numbering of columns, 710 
of girders, 711 








O 


Oberlander, J. T., 833 
Office buildings, 803 
floor plan, 804 
interior, 761 
Office, Salada Tea Co., 759 
Openings, roof, 660 
slabs, 375 
Orchestra floor, 763 
Organic matter, effect, 2 
Ornamental ceilings, 614 
facades, 754 
Outside columns, foundations for, 478 


ie 


p, values for beam and slabs 882 
p’, ratio for given pi, 904 


960 


pa, values in spiral columns, 931 
Paine Furniture Company’s roof, 651 
Painted concrete, 758 
Painting window sash, 703 
Panels. See also Interior panels. 
See also Wall panels. 
. building codes for, 400 
sizes, warehouses, 789 
steel tile construction, 603 
wall vs. interior, 104 
Parabolic distribution stress, 30, 143 
Parallel bars, spacing, 273 
Parapets, 656 
Passive earth pressure, 839 
Pedestals, 486 
pile, 557 
Peerless pile, 557 
Penetration of piles, 543 
Penn Charter Gymnasium ceiling, 607 
- Pent house for elevator shaft, 696 
Pent houses, 678 
Philadelphia Building Code, 218 
columns, 408, 421, 437, 448, 450, 
464 
Pier 6, roof truss, Cristobal, 677 
Piers, strength of brick, 77 
Pilasters, 632 
Piles, 477, 542 
See also Cast-in-place piles 
See also Pre-cast piles 
Bignell pre-cast, 552 
brooming, 545 
caisson, 560 
capacity, cast-in-place, 558 
cast-in-place, 554 
Chenoweth pre-cast, 551 
composite, 555 
concrete, 546 
Cummings pre-cast, 552 
cutting of wood, 544 
driving, 545 
driving pre-cast piles, 550 
Engineering News formulas, 543 
frictional resistance, 543 
Giant pre-cast, 552 
Gilbert pre-cast, 552 
inclined, 545 
independent footings, 516 


INDEX 


Piles, outside bearing walls, 543 
pedestal, 557 
Peerless cast, 557 
penetration, 543 
pre-cast, 546 
raft foundation, 539 
Raymond cast, 554 
retaining walls, 843 
Simplex cast, 555 
spacing, 542 
stresses in pre-cast, 548 
Wellington formulas, 543 
wooden, 543 
Pipes imbedded in concrete, 357 
Pitch of roofs, 649 
of spiral reinforcement, 431 
Plain concrete, columns, 403 
factory, 729 
footings, 480. 
section, direct stress and bending 
on, 169 
general, formula for stress on, 169 
Plan, Pier 6, roof truss, Cristobal, 678 
Plans, building, 707 
flat slab building, 718 
floors, 713 
foundations, 712 
working, 720 
Plaster casts, 757 
finish, 758 
Plastic deformation of concrete, 34. 
Plate anchorage, 163 
Platforms for access, 690 
Plinth, 322 
Point of bending reinforcement, 287 
Points of inflection, 110 
beam and tile floors, 595 
continuous beam, 289 
effect of reinforcement on, 386 
flat slabs, 119, 336 
Poisson’s ratio, 76, 121, 197, 199 
Post, George B. & Son, 750, 800 
Pre-cast piles, 546, 547, 548, 550 
roof trusses, 677 
Pressure on raft foundation, 538 
wall foundations, 841 
Pressures on retaining walls, 836 
Prior, J. H., 872 


INDEX 


Probst, T. E., 64, 67 

Properties of sections, 945 

Proportions for concrete, 4 

Protective covering, reinforcement, 272 
Pull-out tests, 51 

Punching shear, 492 


R 


R, values of, for beam and slab, table, 
205, 880, 881, 882 
diagram, 883 
r, effect of cost of formwork on, 221 
Raft foundation, 538 
Railroad sidings for warehouses, 791 
Ramps for garages, 798 
. circular, 799 
D’ Hume, 801 
Ramp Building Corporation, 797 
Rankine’s theory, earth pressure, 835 
‘Raymond pile, 554 
Recesses on plans, 720 
Rectangular bands, 88 
Rectangular beams, 
see also Beams 
approximate formulas, 207 
balanced design, 202 
compression reinforcement, 232 
economical design, 213 
formulas for design, 203 
review, 206 
plain sections, stress formula, 170 
safe loads, table, 884 
stress distribution, 180 
theory, 127 
table of constants, 880 
Reduction of live load on columns, 452 
for large areas, 453 
Reglets, 658 
Reinforced concrete bearing walls, 634 
definition, 122 
Reinforced concrete, definition, 122 
Reinforcement, 7 
allowable stresses, 8, 879 
anchorage of. See Anchorage. 
angle of bend, 291 
bearing stresses, 54 
bending bars, 355 


961 


Reinforcement, bending diagrams for, 
291 
bond stresses, 10 _ 
circumferential vs. band, 117 
Clinton wire fabric 17 
compression, 50 
corrosion of, 9 
definition of, 7 
deformation in, 28, 72 
deformed bars, 10 
diagrams to determine points of 
bending, 292-297 
Elcannes bars, 14 
expanded metal, 15 
factor of safety, 8 
failure of, 22 
foor plan drawing, 726 
four-way flat slab system, 353 
fulcrum girders, 767 
grade steel, 8 
hard steel, 9 
Havemeyer bars, 13 
identification tags for, 723 
intermediate grade, 9 
Kahn new rib bars, .13 
Kahn trussed bars, 13 
lapping of bars, 413, 
length of inbedment in columns, 462 
lap, 414 
spiral, 433 
metal lath, 18 
minimum, for columns, 413 
for economy, 136 
moisture protection for, 273 
Monotype bars, 14. 
necessity of securing in place, 356 
neutral axis, influence on, 27 
over doors, ete., 634 
per cent, influence of, 25 
pitch. of spiral, 432 
plain vs. deformed, 10 
point of bending in beams and 
slabs, 209, 287, 290 
properties of bars, 12 
properties of bars, steel, 7 
protection of, 9, 272 
ratio in slabs, 211 
T-beams, 894 


962 


Reinforcement, rib lath, 17 
rings, action of, 373 
shapes, 10 
shock, protection from, 10 
shrinkage stresses, 298 
slabs over girders, 211 
Smulski flat slab system, 367 
spacing for aggregates, 273 
splicing of, 61, 265, 267 
steel, brittleness of, 10 — 
structural shapes for columns, 88, 
44] 
supports, 209 
tables of sizes, 942 
temperature, in walls, 855 
temperature stress, 298 
temperature under, 9 
three-way flat slab system, 374 
torsion, resistance to, 94 
triangle mesh, 17 
two layers, 30 
two-way, flat slab system, 352 
vertical spacing, 274 
weight of spiral, 433 
wire, 14, 17 
yield point, significance of, 202 
Relative cost of buildings of different 
materials, 564 
Repeated loads on columns, 77 
Reservoirs. See Vol. III 
Restraint of ends of beam, 275 
slabs at wall, 333 
Retaining walls, 832 
cantilever, 849 
cellular, 869 
drainage, 834 
economy of types, 832 
factor of safety, 842 
failure, 841 
L-shaped, 849 
line of pressure on, 840 
piles under, 843 
pressure on foundation, 846, 848 
sliding of, 838 
T-shaped, 849 
temperature reinforcement, 855 
upright slab, 850 
Rib lath, 19 


INDEX 


Rich mix, economy in rectangular 
beams, 214 
“ economy in slabs, 213 
Rigby, Edward H., 835 
Rigid frame, with arched roof, 667 
roofs, 665 


Ring reinforcement, 370 


Rings, effectiveness as reinforcement, 
352, 373 
splicing, 121 
stress distribution in, 110 
Riplex, 611 
Risers, stairs, 688 
Road construction, 12 
Roadbed, distribution of load on, 75 
Roads and Pavements. See Vol. II 
Robbins, Henry C., 728 
Rock Island R. R., 834 
Roof, air space below, 652 
arched with rigid frame, 667 
cinder fill for, 653 
concrete arches for, 668 
condensation under, 649 
construction, 648 
coverings, classes, 654 
dead loads, 648 
design, 659 
drainage of, 649 
expansion joints in, 653 
foundry for Garfield Smith & Com- 
pany, 666 
girder, Winston-Salem Auditorium, 
663 
insulation of, 650 
long span, 661, 668, 671 
openings in, 660 
Paine Furniture Company, 651 
pitch of, 649 
rigid frame, 665 
sawtooth, 675 
theater, 778 
trusses, concrete, 671 
Grauman’s theater, 674 
pre-cast, 678 
walks on, 659 
wind pressure on, 649 
Rubble masonry, 832 
Runs and platforms, 690 


INDEX 


S 


Safe loads, rectangular beams, 884 
slabs, 886 
Sagar, W. L., 97 
Salada Tea Company Building, 752, 
757, 759 
Sand, organic matter in, 2 
testing, 3 
Sash, dimensions, 700 
standard sizes, 705 
Underwriters’ pivoted, 706 
Sawtooth roofs, 675 
Scagliola, 760 
Scheit, H., 61, 64 
Schmidt, Garden & Martin, 671, 681 
School buildings, 808 
Scuttles to roof, 686 
Sea water, concrete in, 6 
Sections, properties of, 945 
Separate bending sketches for bars, 726 
Settlement, foundations, 472 
unequal, 472 
Shafting inserts in ceiling, 615 
Shapes, reinforcement, 10 
Shear, beams, 143, 244 
cantilevers, 773 
chimneys, 817, 819 
column heads, 913 
definition of, 143 
diagram, moving loads, 258 
drop panels, 914 
external, magnitude of, 144 
footings, 500 
formula for stress, 148 
horizontal, 144, 150 
longitudinal, cause of, 144- 
longitudinal T-beam, 144 
magnitude, solution for, 145 
strap beams, 533 
stress, 148, 148, 151, 244, 346 
vertical, 56, 144, 150 
Shepard & Stearns, 803 
Shock on concrete, 10 
Shrinkage stresses, reinforcement for, 
298 
Sidewalks. See Vol. II 
Simplex pile, 555 


963 


Simplex system reinforcement, 359 
Sinks, Frank F., 362 
Skeleton concrete buildings, 572 
Skylight openings, 660 
Slabs. See also Basement slabs 
See also Flat slabs 
area round bars, 945 
square bars, 945 
arrangement of reinforcement for, 
210 
basement, 618 
bending moment in continuous, 209, 
278 
bending of reinforcing bars for, 210 
breaking load for wide, 72 
cantilever, tests on, 116 
constants, table of, 880, 881 
for selected stresses, 205 
continuous, 208 
cross reinforcement of, 211 
over girders, 211 
design, 205, 208 
constants for, 205, 880, 881 
tables, 886, 888-893 
details and schedule, 716 
on floor plan, 726 
distribution of load, square or ob- 
long, 212 
eccentric loads on, 70 
economy of rich mix, 213 
effective width for concentrated 
load, 69, 70, 72 
example, 579 
formulas for design, 208 
reinforcement, arrangement of, 210 
formula for, 211 
over girders, 211 
safe loads on tables, 886, 888-893 
span of, 209, 277 
steel, points of bending, 209 
ratio, 211 
stresses in, 131 
structural steel beams, 611 
stresses in wide, under 
trated load, 69 
Slabs, supported by four beams, 211 
temperature steel for, 618 
tests, 99, 116 


concen-= 


964 INDEX 
Slabs, thickness, formula for, 208 Spiral columns, formulas, Boston, 
two-way hollow tile and structural Cleveland, and Philadelphia 
steel, 613 Codes 4222) 


Slate shingles, 656 
Slater, W. A., 31, 113 
Sleepers, 624 
Sliding of retaining walls, 838 
Slip of tension bars, 51 
Sloped footings, 490 
Smoke-proof stairways, 686. 
Smulski, Edward, 99, 110, 111, 112, 
116, 224, 233, 367, 784 
Smulski flat slab tests, 104 
Smulski flat slab system, action of, 369 
flat slab sections, 719 
reinforcement, 353, 367 
_ schedule, 727 
Soil, bearing power of, 470 
carrying capacity of, 469 
definitions of varieties, 471 
effect of organic matter on bearing 
power, 470 
unequal settlement, 472 
Southern Engineering Company, 784 
Spacers for reinforcement, 275, 357 
bar, 211 
spiral columns, 433 
Spacing, area bars for various, 945 
columns, 793, 797, 804 
reinforcement. See type in question 
Spalling, 91 
Span length, 277 
Spandrel, beam faced with brick, 746 
beams, 334 
loads, 459 
torsion, 336 
Spatter coat, 736 
Special sash, 705 
Spiral columns, 419 
See also Columns 
description, 419 
design, 431 
details, 482 
economical design, 448 
equivalent area vertical steel, 432 
columns, example, 456 
formulas, authors’ recommendation, 
421 


Chicago Code, 425 
New York Code, 424 

oblong, 430 

octagonal, 429 

ratio length to diameter core, 422 

relative economy, 449 

square, with round spiral, 429 

square spiral not effective, 430 

tests on, 82 
Spiral reinforcement, columns, 160, 419 

cost, 434 

length, 433 

pitch, 431 

relation of pitch 

spiral, 431 

weight, 432 
Spitzer, 79, 93 
Splices in reinforcement, 61 
Splicing, rings, 121 

staggering of, 163 

structural columns, 441 
Spouting of concrete, 357 
Sprinklers, 577 
Stairs, 682, 810, 811 

capacity, 684 

enclosed, 685 

intermediate landings, 690 

landing beam, 691 

layout of, 686 

methods of design, 690 

nosing, 688, 689, 694 

reinforcement, 693 

slope of, 689 

special design, 630 

three-flight, 693 

treads and risers, 688 

two-flight, 690 

winding, 685 
Stairways, smoke proof, 686 
Standard sizes of sash, 705 
Standards for window sash, 699 
Statical limitations, flat slab theory, 194 
Statler Garage, 750, 800, 801 
Steel. See Reinforcement 

bearing plates, 633 


and ratio of 


INDEX 965 


Steel, brackets, 442 
columns, 314 
cylinder foundations, 562 
ratio modulus to that of concrete, 201 
tile, dimensions, 604 
tile floors, 602 
window sash, 699 
Stepped footings, 487 
Stirrups, 36, 38, 96, 242, 251 
action of, 153 
area of vertical, 154, 248 
counterforts, 868 
design tables, 899 
diagonal tension in, 153 
formulas for design, 247, 248 
fulerum girders, 768 
graphical spacing, 252 
held by top bars, 290 
inclined, 243 
limiting spacing, 156, 250 ) 
number of, for uniformly distributed 
loading, table, 900 
spacing depends upon type of 
loading, 253 
spacing for vertical, 154, 247 | 
for inclined, 156, 249 
for uniformly distributed load, 
table, 901 
spacing table, 899 
support for, 290 
theory, 147 
types of, 243 
Stone & Webster, 755 
Stone exterior, 753 
Straight-line formula, 127, 142 
theory, 123 
Strap beams in footings, bending at 
interior columns, 537 
connecting footings, 533 
loading, 534 
pile foundation, 536 
types, 538 
Strength, brick piers, 77 
Stresses, actual and computed, 24, 125, 
128 
allowable unit, table, 879 
allowable, in concrete at supports, 
223 


Stresses, assumptions of distribution 
in rectangular beam, 143 
bearing in hooks, 163 
direct and flexure. See 
stress and bending 
eccentric thrust, caused by. See 
Direct stress and bending 
effect of moduli on, 202 
interdependence between steel and 
concrete, 206 
parabolic distribution of, 30, 
143 
relation to deformation, 387 
rings, distribution in, 110 
steel and concrete compared, 87 
unit, 247 
variation, 127, 143 
wall with counterforts, 865 
Structural steel, and concrete balcony 
cantilevers, 776 
and concrete columns, difficulties of, 
452 
and concrete columns, economy, 452 
and concrete floors, 611 
Structural steel columns, base for, 443 
brackets for, 443 
encased in concrete, details of, 
440, 435 
Support of veneer, 751 
Supports, tension in steel at, 288 
working stresses at, 282 
Surcharge, 638, 837 
Surfaces, coloring of, 737 
floor, 620 
rubbing of concrete, 735 
treatment of concrete, 734 
Suspended ceilings, 19, 6138, 652 
Salada Tea Building, 760 
Suspension action of flat slabs, 387 
Symbols, table of, 128 


Direct 


T 
T-beams, 35 
area of steel in, 135, 221 
compression below flange, 133 
simplified formula, 224 
compression flange, 116, 217 
reinforcement, 225, 230 


966 


INDEX 


T-beams, depth at center governed | Tensile strength of concrete, 21 


by compression, 219 
design, 215 
design diagrams, 894, 895 
tables, 897 
details of design, 223 
diagram of steel ratio, 894 
economical depth, 220 
exact vs. approximate formulas for, 
221 
failure by tension, 35 
fillets, effect on strength, 146 
flange, effective width of, 36 
formulas for design, 216 
simplified, | compression 
flange, 224 
simplified irregular flange, 228 
unsymmetrical flange, 142 
formulas, derivation of, 133 
fulerum girders, 767 
how obtained, 215 
longitudinal shear in, 144 
minimum depth, 135, 220 
minimum depth at support, con- 
tinuous beam, 222 
moment arm for, 221 
moment of resistance, 136 
neutral axis for, 133 
table, 897 
shearing stresses in flanges, 37 
stresses in, 224 
test of, 35 
theory of, 133 
vs. truss, 772 
width of flange, 36, 217 
T-shaped retaining walls, 849 
Tags on reinforcing bars, 723 
Talbot, Arthur N., 28, 35, 47, 76, 80, 
83, 88, 89, 93, 118, 148 
Taper of piles, 542 
Tapered tile, 605 
Tar and gravel roofs, 654 
Taylor, C. Percy, 812 
Temperature in chimneys, 822 
Temperature reinforcement, 
ment walls, 635 
retaining walls, 855 
stresses, reinforcement for, 298 


below 


base- 


Tension, and bending combined on 
rectangular section, 191 
center of, in chimneys, 816 
concrete beam in, 140 
diagonal, 147 
direct, 162 
entire section, subjected to direct, 
190 
members, composition of, 162 
thickness of, 164 
trusses, 770 
Tension reinforcement of columns, 461 
of slab across beams, 146 © 
steel, center of span, 287 
Tension, failure by, of T-beam, 35 
Teredo, 542 


Terra cotta exterior, 753 


Terrazzo floors, 623, 805 
Tests, 20 
beams with compressive reinforce- 
ment, 50 
bend test of bars, 8 
bond of concrete and steel, 51 
cement, 2 
columns with vertical steel, 78 
spiral reinforcement, 82 ¥ 
structural steel, 88 
continuous beams, 115 
cracks in beams, 115 
Deere & Webber Building, 113 
diagonal tension, 38 
flat slab construction, 98 
long columns, 93 
octagonal flat slab, 116 
plain concrete columns, 75 
pull-out of bars, 51 
sand, 3 ¢ 
slabs, 71, 100, 117, 120 
slump, 4 
Smulski system, 104 
T-beams, 36 
Turner-Carter Building, 115 
value of, for structures, 113 
Wenalden Building, 114 
Test pits for ground exploration, 469 
Theater balcony truss, 779 
Theaters, concrete construction, 762 


INDEX 


Theory, flat slab, 194 
least work, 313 
rectangular beams, 

126 
reinforced concrete, 122 

Thompson & Binger, 665, 738 

Thompson, Sanford E., 116, 621, 755 

Three-flight stairs, 693 

Three-way system of reinforcement, 

374 

Thrust and bending moment, 164 
allowable unit stresses, 463 
character of stresses, 463 
circular sections, plain, 172 

reinforced, 179 
columns under, 458 
design of, 460 
condition of loading for maximum 
stress, 187 
diagrams for design: 
part of section in tension, 938, 939 
tension side reinforced, 940, 941 
whole section under compression, 
934-936 
effective section, 179 
general formula, 169 
negative thrust, 190 
part of section under compression, 
191 
whole section under tension, 190 
plain section, 170 
relation between position of eccentric 
thrust and sign of bending 
moment, 166 
reinforced section, general formulas, 
173 
circular, 179 
* one surface intension, 180 
rectangular, 175 
reinforcement in tension face only, 
185 
sign of, governed by bending mo- 
ment, 165 

Tie rods, long arched roofs, 669 
retaining walls, 872 

Ties for facing, 751 
sag in, 163 

Tile floor, construction, 589 


assumptions, 


967 


Tile floor, weight, 590, 597 
Tile insulators, 653 
Tiles, adjustable metal, 605 
gypsum 592 
hollow clay, 590 
metal, 611 
removable, 611 
tapered, 605 
Toe-cantilever walls, 852 
Top bars to hold stirrups, 290 
Topping for hollow tile floors, 593 
for metal tile floors, 608 
pouring, 610 
Toronto tests, 97 
Torsion, cause of, 94 
failure by, 95 
lintel beam, 104 
resistance of concrete in, 94 
spandrel beams, 336 
stresses due to, 94 
ultimate stress, 96 
Traprock, 272 
Trautwine, John C., 840 
Treads, stairs, 686 
Treatment of concrete surfaces, 734 
Triangle mesh, 17 
Truck loads, distribution on road bed, 
75 
Truck-loading platforms, 791 
Trucks, dimensions of, 797 
Trusses, 769 
Turneaure, F. E., 22, 220 
Turner,’ C: P? A;, 359 


Turner-Carter Building, test on, 
115 

Turner Construction Company, 729, 
744 


Two-flight stairs, 690 

Two-way system reinforcement, 362 
flat slab tests, 99 

Types of concrete buildings, 785 
floor construction, 785, 787 


U 


Unbalanced design, rectangular beams, 
132, 203 
Underwriters’ pivoted sash, 706 


908 


Unequal settlement, footings, 537 
raft foundation, 539 

Uniform loading, stirrups for, 253 

Uniformly distributed load, diagonal 
tension with, 156 

United Fireproofing Company, 760, 800 

Universal Portland Cement Company, 
565 


V 


Villadsen Brothers, 665 
Variations in bending moment, 335 
Vertical bars for columns, 434 
Vertical shearing, measure of diagonal 
tension, 241 
spacing, reinforcement, 274 
Voids due to reinforcement, 273 


W 


Walks on roofs, 659 
Wall beams, 379, 382 
general remarks, 385 
Wall bearing construction, 572, 629 
vs. Skeleton construction, 809 
column footing, continuous, 518 
columns, anchorage of steel, 462 
bending of, 458 
deflection of, 460 
footings, 482 
foundations, 841 
panels in flat slab, 333 
panels in flat slab, New York Code, 
400 
Chicago Code, 400 
restraint of flat slabs, 333 
tie rods, 872 
Walls. See also Basement walls 
See also Retaining walls 
basement, 637 
supported at columns, 643 
supported top and bottom, 638 
concrete, 634 
waterproofing of, 637 
Warehouse, column spacing 
construction, 785 
elevators, 791 
floor finish, 791 


INDEX 


Warehouse, live loads, 453, 789 
panel sizes, 789 
wash borings, 469 
Wawrzyniok, O., 61 
Washington Garage, inclined drive- 
way, 799, 800 
Water jet, pile driving, 545 
Water pressure on basement, 619 
Waterproofing walls, 746 
Watertight work, 6, 98 
Water tower, Clark Biscuit Company, 
680 
Watertown Arsenal Tests, 77, 80 
Wayss, 87, 88 
Weather tightness, curtain wall, 382 
Web area of T-beam, 218 
determined by diagonal tension, 
218 
resistance, effect of age on, 47 
Web reinforcement, 242 
action of, 152 
anchorage of, 244 
area and spacing, 248 
cantilevers, 158 i 
continuous beams, 157 
description of, 243 
design of, 251 
proportion of diagonal tension 
resisted by, 153, 245 
tests, 48 
theory, 147 
usefulness of, 157, 242 
Web resistance, effect of horizontal 
steel on, 47 
Wedge-shaped beams, 140 
Weight, clay tile floors, 597 
earth, 833 
of bars, table, 942 
Welded wire fabric, 17 
Wellington formulas for piles, 543 
Wenalden Building Test, 114 
Westergaard, H. M., 196 
Western Newspaper Union Building, 
370 
destruction test, 346 
Width of flange, T-beams, 217 
Wind, chimney loads, 812, 825 
pressure on roofs, 649, 669 


INDEX 969 


Winding stairs, 685 Working stresses, table of, 879 
Window opening, 705 

sash, 699, 701, 703 | Y 

inston-Sal Auditori 663, 762 : : ; 
Bees a8 7a0) 78 eas 3 : ’ | Yield pont effect of reinforcement on, 


Wire reinforcement, 14, 432 Arena 
Wiren, George R., 748 5 is re fin 93 
Withey, M. O., 48, 58, 59, 79, 83, 91, 93 ee a ee 
Young, C. R., 97 
Wood, block floors, 626 3 ee Pie 
floors, 624, 805 Youths’ Companion Building, 740, 793, 
piles, 543 794 
cutting top of, 544 4 
sleepers for floors, 624 
Working plans, 721 Zipprode, R. R., 31 


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